CS 207m: Discrete Structures (Minor) - IIT Bombay
Transcript of CS 207m: Discrete Structures (Minor) - IIT Bombay
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CS 207m: Discrete Structures (Minor)
Instructor : S. Akshay
Jan 05, 2018
Lecture 01 – Introduction
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Logistics
Course hours: Slot 5;Wed 09:30-11:00, Fri 09:30-11:00
Office hours: By prior appointment
Evaluation
I Quizzes: 30%
I Midsem: 25%
I Endsem: 40%
I Other (pop quizzes, participation, assignments): 5%
Course material, references will be posted at
I http://www.cse.iitb.ac.in/~akshayss/teaching.html
I piazza (will be set up by TAs soon)
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Logistics
Course hours: Slot 5;Wed 09:30-11:00, Fri 09:30-11:00
Office hours: By prior appointment
Evaluation
I Quizzes: 30%
I Midsem: 25%
I Endsem: 40%
I Other (pop quizzes, participation, assignments): 5%
Course material, references will be posted at
I http://www.cse.iitb.ac.in/~akshayss/teaching.html
I piazza (will be set up by TAs soon)
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Goal
First things first...
I What are discrete structures?
I Why are we interested in them?
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Course Outline
What we will broadly cover in this course
1. Mathematical reasoning: proofs and structures
2. Counting and combinatorics
3. Elements of graph theory
4. Introduction to abstract algebra & number theory
What we don’t cover
1. Discrete probability
2. Algorithms
3. Logic and Finite automata
4. Details and applications of everything above – rest of your(minor) life!
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Course Outline
What we will broadly cover in this course
1. Mathematical reasoning: proofs and structures
2. Counting and combinatorics
3. Elements of graph theory
4. Introduction to abstract algebra & number theory
What we don’t cover
1. Discrete probability
2. Algorithms
3. Logic and Finite automata
4. Details and applications of everything above – rest of your(minor) life!
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Course Outline
What we will cover in this course
1. Mathematical reasoning: proofs and structures
2. Counting and combinatorics
3. Elements of graph theory
4. Introduction to abstract algebra and number theory
Textbooks
I Discrete Mathematics and its Applications withCombinatorics and Graph Theory, by Kenneth H Rosen.
I Discrete Mathematics by Norman Biggs.
I Introduction to Graph theory by Douglas B West.
I More will be listed on webpage as we go along.
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More lofty aims of the course
1. Introduce mathematical background needed in variousbranches of computer science (and in other sciences!)
2. (New and old) techniques for problem solving: how toattack problems that you have never seen before.
3. To write proofs and convey your ideas formally.
4. To develop skills to read/understand/solve new material inthe future.
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Chapter 1: Proofs and Structures
Outline of next few classes
I Propositions, statements
I What/why of proofs and some generic proof strategies
I Mathematical induction
I Notions and properties of sets, functions, relations
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Propositions
What is a proposition?
I It is raining
I 1 + 1 = 2
I every odd number is a prime
I 267 − 1 is a prime
I (n + 1)(n− 1) = (n2 − 1) for any integer n
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Propositions
What is a proposition?
I It is raining
I 1 + 1 = 2
I every odd number is a prime
I 267 − 1 is a prime
I (n + 1)(n− 1) = (n2 − 1) for any integer n
What is common between these statements?
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Propositions
What is a proposition?
I It is raining
I 1 + 1 = 2
I every odd number is a prime
I 267 − 1 is a prime
I (n + 1)(n− 1) = (n2 − 1) for any integer n
A proposition is a statement that is either true or false (butnot both).
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Propositions
What is a proposition?
I It is raining
I 1 + 1 = 2
I every odd number is a prime
I 267 − 1 is a prime
I (n + 1)(n− 1) = (n2 − 1) for any integer n
A proposition is a statement that is either true or false (butnot both).
Give an example of a statement that is not a proposition.
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Propositions
What is a proposition?
I It is raining
I 1 + 1 = 2
I every odd number is a prime
I 267 − 1 is a prime
I (n + 1)(n− 1) = (n2 − 1) for any integer n
A proposition is a statement that is either true or false (butnot both).
Give an example of a statement that is not a proposition.
I x + 1 = 8
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Propositional calculus
Figure: Aristotle (384 – 322 BCE)
I propositions are statements that are either true or false.
I Just as we use variables x, y, . . . for numbers, we will usevariables p, q, . . . for propositions.
I “if it is raining, it will be wet” : p→ q
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Propositional calculus
Combining propositions
If p, q are propositions, then so are:
I ¬p - not every prime number is odd
I p ∨ q - either it is raining or it will be wet
I p ∧ q - it is raining and it is wet
I p→ q
I p iff q
Can all mathematical statements be written this way?
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Propositional calculus
Combining propositions
If p, q are propositions, then so are:
I ¬p - not every prime number is odd
I p ∨ q - either it is raining or it will be wet
I p ∧ q - it is raining and it is wet
I p→ q
I p iff q
Can all mathematical statements be written this way?
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Propositional calculus
Combining propositions
If p, q are propositions, then so are:
I ¬p - not every prime number is odd
I p ∨ q - either it is raining or it will be wet
I p ∧ q - it is raining and it is wet
I p→ q
I p iff q
Can all mathematical statements be written this way?
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Propositional calculus
Combining propositions
If p, q are propositions, then so are:
I ¬p - not every prime number is odd
I p ∨ q - either it is raining or it will be wet
I p ∧ q - it is raining and it is wet
I p→ q
I p iff q
Can all mathematical statements be written this way?
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Propositional calculus
Combining propositions
If p, q are propositions, then so are:
I ¬p - not every prime number is odd
I p ∨ q - either it is raining or it will be wet
I p ∧ q - it is raining and it is wet
I p→ q
I p iff q
Can all mathematical statements be written this way?
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Predicates and quantifiers
Consider again...
I ∀n
∈ N
(n + 1)(n− 1) = (n2 − 1)
I ∀x, ∃y,
x, y ∈ Z
x = y + 8
I ∀n stands for all values of n in a given domain
I ∃n stands for exists n
I ∈ is the “element of” symbol
I N stands for all natural numbers
I Z stands for all integers
I R, Q, ...
Some propositions are not so easy to “determine”...– e.g., 267 − 1 is not a prime.
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Predicates and quantifiers
Consider again...
I ∀n
∈ N
(n + 1)(n− 1) = (n2 − 1)
I ∀x, ∃y,
x, y ∈ Z
x = y + 8
I ∀n stands for all values of n in a given domain
I ∃n stands for exists n
I ∈ is the “element of” symbol
I N stands for all natural numbers
I Z stands for all integers
I R, Q, ...
Some propositions are not so easy to “determine”...– e.g., 267 − 1 is not a prime.
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Predicates and quantifiers
Consider again...
I ∀n ∈ N (n + 1)(n− 1) = (n2 − 1)
I ∀x, ∃y, x, y ∈ Z x = y + 8
I ∀n stands for all values of n in a given domain
I ∃n stands for exists n
I ∈ is the “element of” symbol
I N stands for all natural numbers
I Z stands for all integers
I R, Q, ...
Some propositions are not so easy to “determine”...– e.g., 267 − 1 is not a prime.
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Theorems and proofs
A theorem is a proposition which can be shown true
Classwork: Prove the following theorems.
1. If m and n are perfect squares of integers, then mn is alsoa perfect square.
2. If 6 is prime, then 62 = 30.
3. Let x be an integer. Then, x is even iff x + x2 − x3 is even.
4. There are infinitely many prime numbers.
5. There exist irrational numbers x, y such that xy is rational.
6. For all n ∈ N, n! ≤ nn.
7. There does not exist a (input-free) C-program which willalways determine whether an arbitrary (input-free)C-program will halt.
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Theorems and proofs
Contrapositive and converse
I The contrapositive of “if A then B” is “if ¬B then ¬A”.
I A statement is logically equivalent to its contrapositive,i.e., it suffices to show one to imply the other.
I To show A iff B, you have to show A implies B andconversely, B implies A.
I Note the difference between contrapositive and converse.
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Proof by contradiction
Theorem 4.: There are infinitely many primes.
Proof by contradiction:
I Suppose there are only finitely many primes, sayp1 < p2 < . . . < pr.
I Let k = (p1 ∗ p2 ∗ . . . ∗ pr) + 1. Then k when divided by anypi has remainder 1. So pi 6 | k for all i ∈ {1, . . . , r}.
I But k > 1 and k is not prime, so k can be written as aproduct of primes (why?)
I Fundamental theorem of arithmetic: any natural number> 1 can be written as a unique product of primes.
I Now let p|k. But p 6∈ {p1, . . . , pr}, so this is a contradiction.
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Proof by contradiction
Theorem 4.: There are infinitely many primes.
Proof by contradiction:
I Suppose there are only finitely many primes, sayp1 < p2 < . . . < pr.
I Let k = (p1 ∗ p2 ∗ . . . ∗ pr) + 1. Then k when divided by anypi has remainder 1. So pi 6 | k for all i ∈ {1, . . . , r}.
I But k > 1 and k is not prime, so k can be written as aproduct of primes (why?)
I Fundamental theorem of arithmetic: any natural number> 1 can be written as a unique product of primes.
I Now let p|k. But p 6∈ {p1, . . . , pr}, so this is a contradiction.
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Proof by contradiction
Theorem 4.: There are infinitely many primes.
Proof by contradiction:
I Suppose there are only finitely many primes, sayp1 < p2 < . . . < pr.
I Let k = (p1 ∗ p2 ∗ . . . ∗ pr) + 1. Then k when divided by anypi has remainder 1. So pi 6 | k for all i ∈ {1, . . . , r}.
I But k > 1 and k is not prime, so k can be written as aproduct of primes (why?)
I Fundamental theorem of arithmetic: any natural number> 1 can be written as a unique product of primes.
I Now let p|k. But p 6∈ {p1, . . . , pr}, so this is a contradiction.
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Proof by contradiction
Theorem 4.: There are infinitely many primes.
Proof by contradiction:
I Suppose there are only finitely many primes, sayp1 < p2 < . . . < pr.
I Let k = (p1 ∗ p2 ∗ . . . ∗ pr) + 1. Then k when divided by anypi has remainder 1. So pi 6 | k for all i ∈ {1, . . . , r}.
I But k > 1 and k is not prime, so k can be written as aproduct of primes (why?)
I Fundamental theorem of arithmetic: any natural number> 1 can be written as a unique product of primes.
I Now let p|k. But p 6∈ {p1, . . . , pr}, so this is a contradiction.
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Proof by contradiction
Theorem 4.: There are infinitely many primes.
Proof by contradiction:
I Suppose there are only finitely many primes, sayp1 < p2 < . . . < pr.
I Let k = (p1 ∗ p2 ∗ . . . ∗ pr) + 1. Then k when divided by anypi has remainder 1. So pi 6 | k for all i ∈ {1, . . . , r}.
I But k > 1 and k is not prime, so k can be written as aproduct of primes (why?)
I Fundamental theorem of arithmetic: any natural number> 1 can be written as a unique product of primes.
I Now let p|k. But p 6∈ {p1, . . . , pr}, so this is a contradiction.
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Proof by contradiction
Theorem 4.: There are infinitely many primes.
Proof by contradiction:
I Suppose there are only finitely many primes, sayp1 < p2 < . . . < pr.
I Let k = (p1 ∗ p2 ∗ . . . ∗ pr) + 1. Then k when divided by anypi has remainder 1. So pi 6 | k for all i ∈ {1, . . . , r}.
I But k > 1 and k is not prime, so k can be written as aproduct of primes (why?)
I Fundamental theorem of arithmetic: any natural number> 1 can be written as a unique product of primes.
I Now let p|k. But p 6∈ {p1, . . . , pr}, so this is a contradiction.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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A Non-constructive proof
Theorem 5.: There exist irrational numbers x and y suchthat xy is rational.
Proof:
I Consider√
2. First show that√
2 is irrational.
I Let x = y =√
2 and consider z =√
2√2.
I Case 1: If z is rational, we are done (why?)
I Case 2: Else z is irrational.
I Then consider z√2 = (
√2√2)√2 = (
√2)2 = 2.
I Thus we have found two irrationals x = z, y =√
2 such thatxy = 2 is rational.
Indeed, note that the above proof is not constructive!
(H.W): Find a constructive proof of this theorem.
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Types of proofs
1. If m and n are perfect squares of integers, then mn is alsoa perfect square.
– Direct proof
2. If 6 is prime, then 62 = 30.
– Vacuous/trivial proof
3. Let x be an integer. Then, x is even iff x + x2 − x3 is even.
– Both directions, by contrapositive (A→ B = ¬B → ¬A)
4. There are infinitely many prime numbers.
– Proof by contradiction
5. There exist irrational numbers x, y such that xy is rational.
– Non-constructive proof
6. For all n ∈ N, n! ≤ nn.
7. There does not exist a (input-free) C-program which willalways determine whether an arbitrary (input-free)C-program will halt.
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Types of proofs
1. If m and n are perfect squares of integers, then mn is alsoa perfect square. – Direct proof
2. If 6 is prime, then 62 = 30. – Vacuous/trivial proof
3. Let x be an integer. Then, x is even iff x + x2 − x3 is even.– Both directions, by contrapositive (A→ B = ¬B → ¬A)
4. There are infinitely many prime numbers.– Proof by contradiction
5. There exist irrational numbers x, y such that xy is rational.– Non-constructive proof
6. For all n ∈ N, n! ≤ nn.
7. There does not exist a (input-free) C-program which willalways determine whether an arbitrary (input-free)C-program will halt.
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