MHZ5355 / MHZ5375
DISCRETE MATHEMATICS
C. P. S. Pathirana
Senior Lecturer
Department of Mathematics & Philosophy of Engineering
The Open University of Sri Lanka
MHZ5355/ MHZ5375
Discrete Mathematics
• Unit 1
: Elementary Number Theory – (05 Sessions)
• Unit 2
: Introduction to Groups and homomorphism – (02 Sessions)
• Unit 3
: Graph Theory – (06 Sessions)
• Unit 4
: Dynamical system and Fractals – (04 Sessions)
• Unit 5
: Theory of Automata – (03 Sessions)
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MHZ5355 / MHZ5375
• Day Schools
•
•
•
•
•
•
Day School 1 & 2 – Elementary Number Theory
Day School 3 & 4 – Introduction to Groups and homomorphism
Day School 5 & 6 – Graph Theory
Day School 7 Dynamical system and Fractals
Day School 8 Theory of Automata
Day School 9 –
Revision
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Unit I- Elementary Number Theory
• Session 1
: Introduction to number theory
• Session 2
: Divisibility and division algorithm
• Session 3
: GCD and LCM
• Session 4
: Properties of prime numbers
• Session 5
: Modular arithmetic
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Session 1 – Introduction to Number Theory
Introduction
• The set of natural numbers - ℕ
The set of natural numbers, also called as positive integers,
consist of counting numbers from 1.
ℕ = {𝟏, 𝟐, 𝟑, 𝟒 … }
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Whole Numbers (𝕎)
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Integer Set (ℤ )
The set {0, ±1, ±2, ±3, … } is called the set
of Integers. It is denoted by ℤ.
i.e.
ℤ = 0, ±1, ±2, ±3, … .
Note – Observe that ℕ is a proper subset
of ℤ. Hence ℕ ⊂ ℤ.
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Rational Numbers (ℚ)
Any number which can be represented as a ratio of
two whole numbers is defined as a Rational Number.
The Rational number set is denoted by ℚ.
i.e.
𝒑
ℚ =
: 𝒑 ∈ ℤ, 𝒒 ∈ ℕ, 𝒒 ≠ 𝟎, 𝒑, 𝒒 𝒅𝒐 𝒏𝒐𝒕 𝒉𝒂𝒗𝒆 𝒄𝒐𝒎𝒎𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓𝒔
𝒒
.
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Irrational Numbers
Any real number which cannot be represented as a
ratio of two whole numbers is defined as an
Irrational Number.
Examples are,
2, 3, 7, 𝑒, 𝜋, …
Real Numbers (ℝ)
The union of the set of rational numbers and the set
of irrational numbers is called the Real Number set.
It is denoted as, ℝ.
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Special Number Set
ℤ+ = 𝑥: 𝑥 ∈ ℤ, 𝑥 > 0
ℤ+
0 = {𝑥: 𝑥 ∈ ℤ, 𝑥 ≥ 0}
ℚ+ = 𝑥: 𝑥 ∈ ℚ, 𝑥 > 0
+
ℚ0 = 𝑥: 𝑥 ∈ ℚ, 𝑥 ≥ 0
ℝ+ = 𝑥: 𝑥 ∈ ℝ, 𝑥 > 0
ℝ+
0 = {𝑥: 𝑥 ∈ ℝ, 𝑥 ≥ 0}
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Imaginary Numbers
The numbers,
−1, −4, −27, …
are called Imaginary Numbers.
Complex Numbers
The set ,
ℂ = {𝑎 + 𝑏𝑖|𝑥, 𝑦 ∈ ℝ, 𝑖 2 = −1}
is defined as the Complex Number set. It is denoted by ℂ.
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ℂ
ℝ
ℚ
ℤ
Imaginary
Numbers
ℕ
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Theorem 1.1
Show that 2 is not a rational.
ProofWe assume that 2 is a rational.
𝑝
That is, 2 =
where 𝑝, 𝑞 ∈ ℤ, 𝑞 ≠ 0
𝑞
and let we consider p and q have no commun factor {gcd 𝑝, 𝑞 = 1}.
𝑝2
Then, 2 = 2 ⇒ 𝑝2 = 2𝑞2 which says 𝑝2 is an even number.
𝑞
2
𝑝 is an even number ⇒ 𝑝 is an even number
Let 𝑝 = 2𝑚 , 𝑚 𝜖 ℤ
⇒ 4𝑚2 =2𝑞2
⇒ 2𝑚2 = 𝑞2 ⇒ which says 𝑞2 is even
⇒ 𝑞 is even
Hence 𝑝 and 𝑞 are same common factor.
This contradicts with the greatest common factor of 𝑝 and 𝑞 is 1.
Hence the assumption is false.
Therefore 2 is not a rational.
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Intervals
Note - Any real number can be uniquely represented in the number line.
Notations
▪ 𝑎, 𝑏 = 𝑥 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑎 < 𝑥 < 𝑏 .
▪ 𝑎, 𝑏 = 𝑥 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑎 ≤ 𝑥 ≤ 𝑏}.
▪ 𝑎, 𝑏 = {𝑥|𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑎 < 𝑥 ≤ 𝑏}.
▪ 𝑎, 𝑏 = 𝑥 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑎 ≤ 𝑥 < 𝑏 .
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1.1 Well Ordering Principle
This states that Every non-empty subset of natural number set has
a least element.
Definition 1.1 (Least Element)
Let 𝐴 be a non empty subset of the set of real numbers. The real
number 𝑙 is called the least element of 𝐴 if, 𝑙 ∈ 𝐴 and for any
𝑥 ∈ 𝐴, 𝑥 ≥ 𝑙.
❖ Least element is an element of A that is smaller than every other
element of A.
❖ For some subset of real numbers it doesn’t have a least element.
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Examples 1.1
i.
0,1 − Least element is 𝟎.
0 is in the set 0,1 and for any 𝑥 ∈ 0,1 , 𝑥 ≥ 0.
ii.
0,1 − Assume that 𝑙 is the least element.
Then 𝑙 ∈ 0,1 . Hence 𝑙 > 0.
𝑙
𝑙
But is in the set and 0 < < 𝑙 .
2
2
This contradict with the assumption.
This set doesn’t have a least element.
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1.2 Archimedean Property
Theorem 1.2
For any 𝑎, 𝑏 ∈ ℕ, there exists a positive integer 𝑛 such that 𝑛𝑎 ≥ 𝑏.
Proof
Assume that the Archimedean property is not true.
Then there exist, some positive Integers 𝑎, 𝑏 such that 𝑛𝑎 < 𝑏 for every 𝑛 ∈ ℕ.
Let 𝑆 = 𝑏 − 𝑛𝑎 𝑛 ∈ ℕ . ⇒ 𝑏 − 𝑛𝑎 > 0 for every 𝑛 ∈ ℕ.
𝑆 is a non-empty subset of ℕ.
By Well Ordering Principle 𝑺 must have a least element. Say 𝒃 − 𝒎𝒂, 𝒎 ∈ ℕ.
Since 𝒎 ∈ ℕ, 𝒎 + 𝟏 ∈ ℕ. ⇒ 𝒃 − (𝒎 + 𝟏)𝒂 ∈ 𝑺 .
But 𝑏 − (𝑚 + 1)𝑎 < 𝑏 − 𝑚𝑎. Contradiction ! (found an element less than least element)
Assumption is false. Therefore The Archimedean Property is true.
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1.3 Mathematical Induction
One way of thinking about mathematical induction is to regard
the statement we are trying to prove as not one proposition, but
a whole sequence of propositions, one for each n. The trick
used in mathematical induction is to prove following steps.
•
•
•
first statement in the sequence, and
prove that if any particular statement is true,
then the one after it is also true.
This enables us to conclude that all the statements are true.
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Example 1.2
Use the Principle of Mathematical Induction to show that, for each 𝑛 ∈ ℕ
1 + 2 + 3 + ⋯+ 𝑛 =
𝑛(𝑛+1)
.
2
(1)
Answer
Let 𝑆 = 𝑛 𝑛 ∈ ℕ, 1 + 2 + 3 + ⋯ + 𝑛 =
For 𝑛 = 1 ⇒ 1 =
Therefore 1 ∈ 𝑆.
Let 𝑘 ∈ 𝑆 ⇒
1(1+1)
2
𝑛 𝑛+1
2
.
equation (1) is true for 𝑛 = 1 .
1 + 2 + 3 + ⋯+ 𝑘 =
𝑘 𝑘+1
.
2
𝑘 𝑘+1
+𝑘+1
2
(𝑘+1)(𝑘+2)
=
2
(𝑘+1)( 𝑘+1 +1)
=
.
2
Add 𝑘 + 1 to both sides, 1 + 2 + 3 + ⋯ + 𝑘 + 𝑘 + 1 =
Hence the equation (1) is true for 𝑛 = 𝑘 + 1 .
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⇒ (𝑘 + 1) ∈ 𝑆.
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By the Principle of Mathematical Induction, for each 𝑛 ∈ ℕ ,
𝑛(𝑛+1)
1 + 2 + 3 + ⋯+ 𝑛 =
.
2
By the Principle of Mathematical Induction, for each 𝑛 ∈ ℕ ,
Example 1.3: 𝟕𝒏 -1 is divisible by 6 for all 𝒏 ≥ 𝟏.
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Let 𝑆 = 𝑛 𝑛 ∈ ℕ, 𝟕𝒏 −1 is divisible by 6 .
For 𝑛 = 1 ⇒ 71 -1=6 = 6(1) is divisible by 6 , is true for 𝑛 = 1 .
Therefore 1 ∈ 𝑆.
We assume true for k,
Let 𝑘 ∈ 𝑆 ⇒ 𝟕𝒌 −1 = 6(𝑘).
let 𝑘 + 1;
𝟕𝒌+𝟏 −1 =
7 𝟕𝒌 − 𝟏
= [(6+1) 𝟕𝒌 ] -1
= 6 𝟕𝒌 + 𝟕𝒌 − 1
= 6 𝟕𝒌 + 6 𝑘 = 6 𝟕𝒌 + 𝑘 .
is divisible by 6
.
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Hence the above statement is true for
𝑛 = 𝑘 + 1 ⇒ (𝑘 + 1) ∈ 𝑆.
By the Principle of Mathematical Induction, for each 𝑛 ∈ ℕ ,
𝟕𝒏 -1 is divisible by 6 for all 𝒏 ≥ 𝟏.
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