1
CSCI 2110 Discrete Mathematics Tutorial 10
Chin
2
About me
• Name: Chin• Office: SHB 117• Email: [email protected]• Office Hour: Friday 10:00 – 12:00– Outside office hour? Email me first
3
Proposition
• A statement that can be true or false, but not both
• Examples– I hate discrete maths– The quizzes are difficult– 1 + 1 = 3– All primes are odd– No one likes this tutorial
4
Predicate
• A function that maps elements from a domain to {true, false}
• A claim with variables• Examples– ocamp(x) := x went to ocamp– attend(y) := y attends 2110 lecture– odd(a, b) := a is odd and b is odd– compare(x, y) := x < y– female(p) := p is female
5
Domain of a predicate
• A set that contains the values a variable can take
• attend(y) := y attends 2110 lecture• Domain can be the set of 2110 students– then attend(Chin) is undefined (neither true, nor false)– because Chin is not in the domain
• Sometimes, we omit the domain when the context is clear
6
Quantifiers
• for ∀ll, – for any, for every, for each, given any, …– P(x) := x goes to tutorial– ∀x, P(x)
• there ∃xists, – (for) some, at least one, (for) one, …– P(x) := x teaches tutorial– ∃x, P(x)
7
Domain is important
• Let P be a predicate from the domain D– P(x) := x loves discrete maths
• Is it true that ∀x ∈ D, P(x)?
• When D is the set of– 2110 tutors, yes– 2110 students, no
8
Negation
• De Morgan’s laws for first order logic:
• ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x)
• ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)
9
De Morgan’s laws
• Let P(x) := x is female• ∀x, P(x) means everyone is female
• Which one is its negation, i.e. ¬(∀x, P(x)) ?– No one is female, i.e. ¬(∃x, P(x))– Everyone is not female, i.e. (∀x, ¬P(x))– Someone is female, i.e. ∃x, P(x)– Someone is not female, i.e. ∃x, ¬P(x)
10
Express the following using predicates
• Every one goes to lectures
• Let attend(x) := x goes to lectures• ∀x, attend(x)
• What is its negation?
11
More quantifiers
• Every one goes to some lectures
• Let attend(x, y) := x attend lecture y• ∀x, ∃y attend(x, y)
• What is its negation?
12
More quantifiers
• What is the negation of ∀x, ∃y attend(x, y)?
• ¬(∀x, ∃y attend(x, y))• ∃x, ¬(∃y attend(x, y))• ∃x, ∀y ¬attend(x, y)
• Some students don’t attend any lectures
13
Order of quantifiers are important
• Every one goes to some lectures• Let attend(x, y) := x attend lecture y
• ∀x, ∃y attend(x, y)– Everyone goes to some lectures– Alice can go to first lecture, Bob can go to the second
• ∃y, ∀x attend(x, y)– There exist some lectures that everyone attends– Everyone goes to the same lectures
14
More quantifiers
• Every year, someone goes to every lecture
• attend(x, y, z) := x goes to lecture y in year z• ∀z, ∃x, ∀y attend(x, y, z)
• What is its negation?
15
More quantifiers
• What is the negation of ∀z, ∃x, ∀y, attend(x, y, z)?
• ¬(∀z, ∃x, ∀y, attend(x, y, z))• ∃z, ¬(∃x, ∀y, attend(x, y, z))• ∃z, ∀x, ¬(∀y, attend(x, y, z))• ∃z, ∀x, ∃y, ¬attend(x, y, z)
• In some years, every student does not go to some lectures.
16
More and more quantifiers
• For every d, there exists x and y, so that for every f, f(x) – f(y) < d
• Let P(f, x, y, d) := f(x) – f(y) < d• ∀d, ∃x, ∃y, ∀f, P(f, x, y, d)
• What is its negation?
17
More and more quantifiers
• What is the negation of– ∀d, ∃x, ∃y, ∀f, P(f, x, y, d)
• ¬(∀d, ∃x, ∃y, ∀f, P(f, x, y, d))• ∃d, ¬(∃x, ∃y, ∀f, P(f, x, y, d))• ∃d, ∀x, ¬(∃y, ∀f, P(f, x, y, d))• ∃d, ∀x, ∀y, ¬(∀f, P(f, x, y, d))• ∃d, ∀x, ∀y, ∃f, ¬P(f, x, y, d)
18
Statements in First Order Logic
• Every odd prime can be written as a sum of the square of two integers
• What are the predicates?– Let P(p) := p is an odd prime and can be written as a
sum of the square of two integers– ∀p, P(p). Correct, but …
• In the quiz and exam, you have to break predicates down whenever it is possible
19
Mathematical Statements
• ∀p, P(p)– P(p) := p is an odd prime and can be written as a
sum of the square of two integers
• P(p) := oddprime(p) ∧ sumoftwo(p)
• Can we do more?
20
Mathematical Statements
• P(p) := oddprime(p) ∧ sumoftwo(p)• oddprime(p) := odd(p) ∧ prime(p)– odd(p) := ∃k , ∈ℤ p = 2k + 1– prime(p) := (p > 1) ∧ (∀a ,∈ℤ ∀b , ∈ℤ ((a > 1) (∧ b > 1) → ab ≠ p))
21
Mathematical Statements
• P(p) := oddprime(p) ∧ sumoftwo(p)– sumoftwo(p) := p = x2 + y2 for some integers x, y
• sumoftwo(p) := p = x2 + y2 for some integers x, y• sumoftwo(p) := ∃x , ∈ℤ ∃y , ∈ℤ p = x2 + y2
22
Mathematical Statements
• P(p) := oddprime(p) ∧ sumoftwo(p)• oddprime(p) := odd(p) ∧ prime(p)– odd(p) := ∃k , ∈ℤ p = 2k + 1– prime(p) := (p > 1) ∧ (∀a ,∈ℤ ∀b , ∈ℤ ((a > 1) (∧ b > 1) → ab ≠ p))
• sumoftwo(p) := ∃x , ∈ℤ ∃y , ∈ℤ p = x2 + y2
• ∀p, P(p)
23
Method of proofs
• Direct proof• Proof by contradiction• Proof by cases
24
Direct Proof
• Let a be an even number and b be an odd. Show that ab is an even number
• Write a = 2s, and b = 2t + 1 for some integers s and t
• Then ab = 2s(2t + 1) = 4st + 2s = 2(2st + s)• Therefore ab is an even number because 2st+s
is an integer.
cannot use the same letter
25
Proof by contradiction
• Show that if 2x + 45 < 85, then x < 20
• Suppose x ≥ 20• 2x ≥ 40• 2x + 40 ≥ 85• Contradiction!
26
Proof by contradiction
• Show that 3 is not rational.
• How to say 3 is a rational number?– 3 = p/q for some integers p and q
√
√√
27
Proof by contradiction
• 3 = p/q for some integers p and q
• We choose p and q so that have no common factors that are greater than 1.– 12/64 → 3/16– You can always reduce a fraction
• If we can show p and q have 3 as their common factor, done. But how?
√
28
Proof by contradiction
• 3 = p/q for some integers p and q– Where p and q have no common factors that are > 1
• 3q2 = p2
• If p2 = 3s, p = 3t for some integer t– Prove by contradiction
√
29
Proof by contradiction
• If p2 = 3s for some integer s, p = 3t for some integer t– Prove by contradiction
• Suppose p ≠ 3t for any t– Case 1: p = 3t + 1 for some t– Case 2: p = 3t + 2 for some t
30
Proof by contradiction
• If p2 = 3s for some integer s, p = 3t for some integer t
• Suppose p2 = 3s for some integer s but p ≠ 3t for any t– Case 1:
p = 3t + 1, p2 = 9t2 + 6t + 1 = 3(3t2 + 2t) + 1p2 ≠ 3s for any s
– Case 2:p = 3t + 2, p2 = 9t2 + 12t + 4 = 3(3t2 + 4t + 1) + 1p2 ≠ 3s for any s
• Contradiction.
31
Proof by contradiction
• 3q2 = p2
– If p2 = 3s for some integer s, p = 3t for some integer t
• p = 3t for some integer t
• 3q2 = p2 = 9t– q2 = 3t– If p2 = 3t for some integer t, p = 3u for some
integer u• q = 3u for some integer u
32
Proof by contradiction
• p = 3t for some integer t• q = 3u for some integer u
• So p and q have 3 as their common factor.
• Contradiction because we begin with the fact that p and q do not have common factor > 1.
33
Proof by contradiction
• Suppose 100 students took a quiz and the mean is 80 (out of 100). Show that at least 50 students score greater than 60.
• What is the negation of at least 50 students score > 60?– At most 49 students score > 60– (or at least 51 score ≤ 60)
34
Proof by contradiction
• What is the highest mean we can get?• The 49 students who score > 60 get 100• The rest of the 51 students who score ≤ 60 get 60
• Then the mean is (# people who score 100) × 100 + (# people who score 60) × 60
= (49 × 100 + 51 × 60) / 100= (4900 + 3060) / 100= 79.6 < 80• Contradiction!
(total number of students)
35
Proof by cases
• Prove that max(x, y) + min(x, y) = x + y.
• If y > x– max(x, y) = y and min(x, y) = x– max(x, y) + min(x, y) = y + x = x + y
• If y ≤ x– max(x, y) = x and min(x, y) = y– max(x, y) + min(x, y) = x + y
• done, since either x < y or y <= x must be true.
36
Proof by cases
• Suppose x > 0 and y > 0.• Prove that (|x + y| - |x – y|)/2 = min(x, y)
• If x > y– min(x, y) = y– |x + y| = x + y , |x – y| = x – y– |x + y| - |x – y| = (x + y) – (x – y) = 2y– (|x + y| - |x – y|)/2 = y = min(x, y)
37
Proof by cases
• Suppose x > 0 and y > 0.• Prove that (|x + y| - |x – y|)/2 = min(x, y)
• If x ≤ y– min(x, y) = x– |x + y| = x + y , |x – y| = y – x– (because |a| = -a if a ≤ 0, and (x – y) ≤ 0)– |x + y| - |x – y| = (x + y) – (y – x) = 2x– (|x + y| - |x – y|)/2 = x = min(x, y)
38
End
• Questions?
Top Related