Intro to Mathematical Reasoning via Discrete Mathematics - - PowerPoint PPT Presentation
Intro to Mathematical Reasoning via Discrete Mathematics CMSC-37115 Instructor: Laszlo Babai University of Chicago Week 1, Tuesday, September 29, 2020 CMSC-37115 Mathematical Reasoning Sets, functions, numbers Sets of numbers: N = { 1 , 2
CMSC-37115 Mathematical Reasoning
CMSC-37115 Mathematical Reasoning
Two operations: addition, multiplication Identities a + b = b + a commutativity of addition ab = ba commutativity of multiplication (a + b) + c = a + (b + c) associativity of addition (ab)c = a(bc) associativity of multiplication An identity that connects the two operations?
CMSC-37115 Mathematical Reasoning
Two operations: addition, multiplication Identities a + b = b + a commutativity of addition ab = ba commutativity of multiplication (a + b) + c = a + (b + c) associativity of addition (ab)c = a(bc) associativity of multiplication An identity that connects the two operations? a(b + c) = ab + ac distributivity
CMSC-37115 Mathematical Reasoning
Ordering (for Z, R): a ≥ b
CMSC-37115 Mathematical Reasoning
Ordering (for Z, R): a ≥ b Most important property of ordering: If a ≥ b and b ≥ c then a ≥ c
CMSC-37115 Mathematical Reasoning
Ordering (for Z, R): a ≥ b Most important property of ordering: If a ≥ b and b ≥ c then a ≥ c Notation (a ≥ b) ∧ (b ≥ c) ⇒ (a ≥ c) What is this property called?
CMSC-37115 Mathematical Reasoning
Ordering (for Z, R): a ≥ b Most important property of ordering: If a ≥ b and b ≥ c then a ≥ c Notation (a ≥ b) ∧ (b ≥ c) ⇒ (a ≥ c) What is this property called? transitivity of the “≥” relation
CMSC-37115 Mathematical Reasoning
Ordering (for Z, R): a ≥ b Most important property of ordering: If a ≥ b and b ≥ c then a ≥ c Notation (a ≥ b) ∧ (b ≥ c) ⇒ (a ≥ c) What is this property called? transitivity of the “≥” relation Relation between ordering and addition?
CMSC-37115 Mathematical Reasoning
Ordering (for Z, R): a ≥ b Most important property of ordering: If a ≥ b and b ≥ c then a ≥ c Notation (a ≥ b) ∧ (b ≥ c) ⇒ (a ≥ c) What is this property called? transitivity of the “≥” relation Relation between ordering and addition? (a ≥ b) ⇒ (a + c ≥ b + c)
CMSC-37115 Mathematical Reasoning
Focus on Z (for a while) Central concept: divisibility “8 divides 48” and “8 does not divide 28” Notation 8 | 48, 8 ∤ 28 Terminology. All of the following statements are synonymous: 8 divides 48 8 is a divisor of 48 48 is a multiple of 8 48 is divisible by 8 Their common abbreviation is 8 | 48. More examples: 9 | 63, 9 | −63, 7 | 63, 37 | 999
CMSC-37115 Mathematical Reasoning
Definition When do we say that a | b ?
CMSC-37115 Mathematical Reasoning
Definition When do we say that a | b ? a | b if there exists x such that ax = b Notation (∃x)(ax = b) We call such an x a witness (to the validity of the satement (∃x)(ax = b) ∃ – existential quantifier “there exists . . . such that”
CMSC-37115 Mathematical Reasoning
Definition When do we say that a | b ? a | b if there exists x such that ax = b Notation (∃x)(ax = b) We call such an x a witness (to the validity of the satement (∃x)(ax = b) ∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” (∀a, b)(a + b = b + a)
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all”
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1)
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1) False: a = 0 is a counterexample: (∄x)(0 · x = 1)
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1) False: a = 0 is a counterexample: (∄x)(0 · x = 1) Tue or false? (∀a)(a 0 ⇒ (∃x)(ax = 1))
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1) False: a = 0 is a counterexample: (∄x)(0 · x = 1) Tue or false? (∀a)(a 0 ⇒ (∃x)(ax = 1)) Depends of the universe over which the quantifiers range
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1) False: a = 0 is a counterexample: (∄x)(0 · x = 1) Tue or false? (∀a)(a 0 ⇒ (∃x)(ax = 1)) Depends of the universe over which the quantifiers range True if the universe is R, false if the universe is Z
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1) False: a = 0 is a counterexample: (∄x)(0 · x = 1) Tue or false? (∀a)(a 0 ⇒ (∃x)(ax = 1)) Depends of the universe over which the quantifiers range True if the universe is R, false if the universe is Z (∀a ∈ R)((a 0) ⇒ (∃x ∈ R)(ax = 1)) TRUE (∀a ∈ Z)((a 0) ⇒ (∃x ∈ Z)(ax = 1)) FALSE
CMSC-37115 Mathematical Reasoning
∃ – existential quantifier “there exists . . . such that” ∀ – universal quantifier “for all” True or false? (∀a)(∃x)(ax = 1) False: a = 0 is a counterexample: (∄x)(0 · x = 1) Tue or false? (∀a)(a 0 ⇒ (∃x)(ax = 1)) Depends of the universe over which the quantifiers range True if the universe is R, false if the universe is Z (∀a ∈ R)((a 0) ⇒ (∃x ∈ R)(ax = 1)) TRUE (∀a ∈ Z)((a 0) ⇒ (∃x ∈ Z)(ax = 1)) FALSE Mixed universes (∀a ∈ Z)((a 0) ⇒ (∃x ∈ R)(ax = 1)) TRUE
CMSC-37115 Mathematical Reasoning
Let us fix Z as our universe. How do we prove that this statement is false: (∀a)(a 0 ⇒ (∃x)(ax = 1)) We need to find a counterexample, i.e., need to show (∃a)(a 0 (∃x)(ax = 1)) What would be such an a?
CMSC-37115 Mathematical Reasoning
Let us fix Z as our universe. How do we prove that this statement is false: (∀a)(a 0 ⇒ (∃x)(ax = 1)) We need to find a counterexample, i.e., need to show (∃a)(a 0 (∃x)(ax = 1)) What would be such an a? a = 2, for instance. (2 0) (∃x)(ax = 1)) When is an inference A ⇒ B false? If A is true but B is false
CMSC-37115 Mathematical Reasoning
Let us fix Z as our universe. How do we prove that this statement is false: (∀a)(a 0 ⇒ (∃x)(ax = 1)) We need to find a counterexample, i.e., need to show (∃a)(a 0 (∃x)(ax = 1)) What would be such an a? a = 2, for instance. (2 0) (∃x)(ax = 1)) When is an inference A ⇒ B false? If A is true but B is false In our case, A is true (2 0); we need to show that B is false, i.e., (∄x)(2x = 1), i.e., (∀x)(2x 1). Which is
Mathematical Reasoning
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion?
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion? (∃z)(az = c)
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion? (∃z)(az = c) Let us guess z:
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion? (∃z)(az = c) Let us guess z: z = xy. Checking: need to show that az = c, i.e., axy = c. But c = by = (ax)y = a(xy) = az.
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion? (∃z)(az = c) Let us guess z: z = xy. Checking: need to show that az = c, i.e., axy = c. But c = by = (ax)y = a(xy) = az.
Associtivity of multiplication
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion? (∃z)(az = c) Let us guess z: z = xy. Checking: need to show that az = c, i.e., axy = c. But c = by = (ax)y = a(xy) = az.
Associtivity of multiplication ⇒ transitivity of divisibility
CMSC-37115 Mathematical Reasoning
“a divides b” a | b e.g., 8 | 24 Recall DEF: a | b if (∃x)(ax = b) A little theorem: (∀a, b, c)((a | b) ∧ (b | c) ⇒ (a | c)) the “divisibility” relation is transitive Proof. Assumptions: (∃x)(ax = b) and (∃y)(by = c) What is the desired conclusion? (∃z)(az = c) Let us guess z: z = xy. Checking: need to show that az = c, i.e., axy = c. But c = by = (ax)y = a(xy) = az.
Associtivity of multiplication ⇒ transitivity of divisibility basic arithmetic ր property of divisibility
CMSC-37115 Mathematical Reasoning
Proposition (a little theorem): (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a + b)) DO (due before tomorrow’s problem session) Prove this result State, what property of basic arithmetic did you use Same with subtraction: (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a − b))
CMSC-37115 Mathematical Reasoning
Proposition (a little theorem): (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a + b)) DO (due before tomorrow’s problem session) Prove this result State, what property of basic arithmetic did you use Same with subtraction: (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a − b)) Which numbers are divisible by 2?
CMSC-37115 Mathematical Reasoning
Proposition (a little theorem): (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a + b)) DO (due before tomorrow’s problem session) Prove this result State, what property of basic arithmetic did you use Same with subtraction: (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a − b)) Which numbers are divisible by 2? (∀x)(2 | x ⇐⇒ x is even)
CMSC-37115 Mathematical Reasoning
Proposition (a little theorem): (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a + b)) DO (due before tomorrow’s problem session) Prove this result State, what property of basic arithmetic did you use Same with subtraction: (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a − b)) Which numbers are divisible by 2? (∀x)(2 | x ⇐⇒ x is even) Which numbers are divisible by 1?
CMSC-37115 Mathematical Reasoning
Proposition (a little theorem): (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a + b)) DO (due before tomorrow’s problem session) Prove this result State, what property of basic arithmetic did you use Same with subtraction: (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a − b)) Which numbers are divisible by 2? (∀x)(2 | x ⇐⇒ x is even) Which numbers are divisible by 1? All numbers: (∀a)(1 | a). Why? Because (∀a)(∃x)(1 · x = a), namely, x =
CMSC-37115 Mathematical Reasoning
Proposition (a little theorem): (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a + b)) DO (due before tomorrow’s problem session) Prove this result State, what property of basic arithmetic did you use Same with subtraction: (∀a, b, d)((d | a) ∧ (d | b) ⇒ (d | a − b)) Which numbers are divisible by 2? (∀x)(2 | x ⇐⇒ x is even) Which numbers are divisible by 1? All numbers: (∀a)(1 | a). Why? Because (∀a)(∃x)(1 · x = a), namely, x = a.
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b)
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0?
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ?
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ? Yes, because (∃x)(0 · x = 0) What is such an x ?
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ? Yes, because (∃x)(0 · x = 0) What is such an x ? x = 17
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ? Yes, because (∃x)(0 · x = 0) What is such an x ? x = 17 (or any other number)
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ? Yes, because (∃x)(0 · x = 0) What is such an x ? x = 17 (or any other number) Can we divide by zero ???
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ? Yes, because (∃x)(0 · x = 0) What is such an x ? x = 17 (or any other number) Can we divide by zero ???
does NOT involve division, only multiplication
CMSC-37115 Mathematical Reasoning
DEF: a | b if (∃x)(ax = b) Which numbers are divisible by 0? Is 0 | 0 ? Yes, because (∃x)(0 · x = 0) What is such an x ? x = 17 (or any other number) Can we divide by zero ???
does NOT involve division, only multiplication
i.e., a | b, a | −b, −a | b, −a | −b
(b) Find all divisors of the number 1. (c) What is the number of divisors of a prime number?
CMSC-37115 Mathematical Reasoning
The universe is still Z. DEF a is congruent to b modulo m if m | a − b. Notation: a ≡ b (mod m) Examples: 5 ≡ 47 (mod 7) b/c 7 | 47 − 5 = 42 = 7 · 6 −3 ≡ 12 (mod 5) b/c 5 | 12 − (−3) = 15 = 5 · 3 Suppose March 2 is Wednesday. What other days of March are Wednesdays?
CMSC-37115 Mathematical Reasoning
The universe is still Z. DEF a is congruent to b modulo m if m | a − b. Notation: a ≡ b (mod m) Examples: 5 ≡ 47 (mod 7) b/c 7 | 47 − 5 = 42 = 7 · 6 −3 ≡ 12 (mod 5) b/c 5 | 12 − (−3) = 15 = 5 · 3 Suppose March 2 is Wednesday. What other days of March are Wednesdays? March 2, 9, 16, 23, 30.
CMSC-37115 Mathematical Reasoning
The universe is still Z. DEF a is congruent to b modulo m if m | a − b. Notation: a ≡ b (mod m) Examples: 5 ≡ 47 (mod 7) b/c 7 | 47 − 5 = 42 = 7 · 6 −3 ≡ 12 (mod 5) b/c 5 | 12 − (−3) = 15 = 5 · 3 Suppose March 2 is Wednesday. What other days of March are Wednesdays? March 2, 9, 16, 23, 30. Day x of March is Wednesday if and only if x ≡
CMSC-37115 Mathematical Reasoning
The universe is still Z. DEF a is congruent to b modulo m if m | a − b. Notation: a ≡ b (mod m) Examples: 5 ≡ 47 (mod 7) b/c 7 | 47 − 5 = 42 = 7 · 6 −3 ≡ 12 (mod 5) b/c 5 | 12 − (−3) = 15 = 5 · 3 Suppose March 2 is Wednesday. What other days of March are Wednesdays? March 2, 9, 16, 23, 30. Day x of March is Wednesday if and only if x ≡ 2 (mod 7) DO (by tomorrow’s problem session) Find all x such that x2 ≡ 4 (mod x). Prove that you found all solutions. DO (by tomorrow’s problem session) Find x, y such that x ≡ 1 (mod 5) and y ≡ 1 (mod 7) and x + y ≡ 1 (mod 9). Explain how you found them. (Trial and error is ok, but try to be smart about your trials.)
CMSC-37115 Mathematical Reasoning
The universe is still Z. DEF a is congruent to b modulo m if m | a − b. Notation: a ≡ b (mod m) DO (by tomorrow’s problem session) Prove that for any fixed value m, “congruence modulo m” is a transitive relation. Formally: (∀a, b, c, m)(((a ≡ b (mod m)) ∧ (b ≡ c (mod m))) ⇒ (a ≡ c (mod m))) (Note that the modulus is the same in the three congruences.) State, what property of divisibility you are using.
CMSC-37115 Mathematical Reasoning
DO (by tomorrow’s problem session) Prove: If a ≡ b (mod m) then (∀c)(ac ≡ bc (mod m)). State what properties of basic arithmetic and of divisibility you are using. Note about omitting universal quantifiers. A completely formal statement would involve universal quantification of the variables a and b: (∀a, b, c)((a ≡ b (mod m)) ⇒ (ac ≡ bc (mod m))). The convention is that if some variables are not bound by a quantifier (these are called “free variables”), then the meaning is that they are universally quantified. For instance, all variables in our identities are meant to be universally quantified: (∀a, b, c)((a + b)c = ac + bc)
CMSC-37115 Mathematical Reasoning
DO (by tomorrow’s problem session) Suppose a ≡ x (mod m) and b ≡ y (mod m). Prove: (a) a + b ≡ x + y (mod m) (b) ab ≡ xy (mod m) State what facts you are using. These facts are not necessarily about basic arithmetic and divisibility. You may also use facts about congruences that we (you) have already proved.
statement means (∀a, b, x, y, m)(((a ≡ x (mod m)) ∧ (b ≡ y (mod m))) ⇒ ((a + b ≡ x + y (mod m)) ∧ (ab ≡ xy (mod m))))
CMSC-37115 Mathematical Reasoning