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Intro to Mathematical Reasoning via Discrete Mathematics CMSC-37115 Instructor: Laszlo Babai University of Chicago Week 2, Tuesday, October 6, 2020 CMSC-37115 Mathematical Reasoning Functions f : A B assigns value f ( a ) B to each

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Intro to Mathematical Reasoning via Discrete Mathematics

CMSC-37115 Instructor: Laszlo Babai University of Chicago Week 2, Tuesday, October 6, 2020

CMSC-37115 Mathematical Reasoning

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Functions

f : A → B assigns value f(a) ∈ B to each a ∈ A domain: set A codomain: set B

CMSC-37115 Mathematical Reasoning

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Functions

f : A → B assigns value f(a) ∈ B to each a ∈ A domain: set A codomain: set B (∀a ∈ A)(∃!b ∈ B)(f(a) = b)

CMSC-37115 Mathematical Reasoning

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Functions

f : A → B assigns value f(a) ∈ B to each a ∈ A domain: set A codomain: set B range: range(f) = {f(a) | a ∈ A} values actually taken

CMSC-37115 Mathematical Reasoning

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Functions

f : A → B assigns value f(a) ∈ B to each a ∈ A domain: set A codomain: set B range: range(f) = {f(a) | a ∈ A} values actually taken range(f) ⊆ B

CMSC-37115 Mathematical Reasoning

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Functions

f : A → B assigns value f(a) ∈ B to each a ∈ A domain: set A codomain: set B range: range(f) = {f(a) | a ∈ A} values actually taken

Example: A = {Alabama, Alaska, Arizona, Arkansas, California, . . . , Wisconsin, Wyoming} B = {3, 4, . . . , 538}

Table: el(x): number of electors from state x

x AL AK AZ AR CA CO CT DE DC FL GA . . . WI WY el(x) 9 3 11 6 55 9 7 3 3 29 16 10 3

range(el) = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 29, 38, 55}

CMSC-37115 Mathematical Reasoning

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Functions

f : A → B assigns value f(a) ∈ B to each a ∈ A domain: set A codomain: set B range: range(f) = {f(a) | a ∈ A} values actually taken

Example: A = {Alabama, Alaska, Arizona, Arkansas, California, . . . , Wisconsin, Wyoming} B = {3, 4, . . . , 538}

Table: el(x): number of electors from state x

x AL AK AZ AR CA CO CT DE DC FL GA . . . WI WY el(x) 9 3 11 6 55 9 7 3 3 29 16 10 3

range(el) = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 29, 38, 55} | range(el)| = 19

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection Notation: BA set of A → B functions f ∈ BA injective if there are no collisions, i.e., (∀u, v ∈ A)(f(u) = f(v) ⇒ u = v) f ∈ BA surjective if range(f) = B, i.e., (∀z ∈ B)(∃u ∈ A)(f(u) = z)

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection Notation: BA set of A → B functions f ∈ BA injective if there are no collisions, i.e., (∀u, v ∈ A)(f(u) = f(v) ⇒ u = v) f ∈ BA surjective if range(f) = B, i.e., (∀z ∈ B)(∃u ∈ A)(f(u) = z) f ∈ BA bijective if injective and surjective, i.e., (∀z ∈ B)(∃!u ∈ A)(f(u) = z)

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection Notation: BA set of A → B functions f ∈ BA injective if there are no collisions, i.e., (∀u, v ∈ A)(f(u) = f(v) ⇒ u = v) f ∈ BA surjective if range(f) = B, i.e., (∀z ∈ B)(∃u ∈ A)(f(u) = z) f ∈ BA bijective if injective and surjective, i.e., (∀z ∈ B)(∃!u ∈ A)(f(u) = z) existence of u means

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection Notation: BA set of A → B functions f ∈ BA injective if there are no collisions, i.e., (∀u, v ∈ A)(f(u) = f(v) ⇒ u = v) f ∈ BA surjective if range(f) = B, i.e., (∀z ∈ B)(∃u ∈ A)(f(u) = z) f ∈ BA bijective if injective and surjective, i.e., (∀z ∈ B)(∃!u ∈ A)(f(u) = z) existence of u means surjectivity

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection Notation: BA set of A → B functions f ∈ BA injective if there are no collisions, i.e., (∀u, v ∈ A)(f(u) = f(v) ⇒ u = v) f ∈ BA surjective if range(f) = B, i.e., (∀z ∈ B)(∃u ∈ A)(f(u) = z) f ∈ BA bijective if injective and surjective, i.e., (∀z ∈ B)(∃!u ∈ A)(f(u) = z) existence of u means surjectivity uniqueness of u means injectivity non-uniqueness means collision: f(u1) = f(u2) = z

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection

Notation: f : A → B (domain → codomain) if f(a) = b then write f : a → b (a maps to b) L

A

T EX a \mapsto b

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection

Notation: f : A → B (domain → codomain) if f(a) = b then write f : a → b (a maps to b) L

A

T EX a \mapsto b Example: f(x) = x2 −3 → 9, √ 2 → 2

CMSC-37115 Mathematical Reasoning

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Injection, surjection, bijection

Notation: f : A → B (domain → codomain) if f(a) = b then write f : a → b (a maps to b) L

A

T EX a \mapsto b Example: f(x) = x2 −3 → 9, √ 2 → 2 Terminology function = map or mapping injective function = injective map = injection surjective function = surjective map = surjection bijective function = bijective map = bijection

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B|

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B}

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B} Proposition (little theorem) |A + B| ≤ |A| · |B|

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B} Proposition (little theorem) |A + B| ≤ |A| · |B| Proof via surjection We know: |A × B| = |A| · |B|. Goal: Find surjection f : A × B → A + B.

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B} Proposition (little theorem) |A + B| ≤ |A| · |B| Proof via surjection We know: |A × B| = |A| · |B|. Goal: Find surjection f : A × B → A + B. So for (a, b) ∈ A × B, need to define f(a, b)

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B} Proposition (little theorem) |A + B| ≤ |A| · |B| Proof via surjection We know: |A × B| = |A| · |B|. Goal: Find surjection f : A × B → A + B. So for (a, b) ∈ A × B, need to define f(a, b) CHAT!

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B} Proposition (little theorem) |A + B| ≤ |A| · |B| Proof via surjection We know: |A × B| = |A| · |B|. Goal: Find surjection f : A × B → A + B. So for (a, b) ∈ A × B, need to define f(a, b) CHAT! f(a, b) := a + b

CMSC-37115 Mathematical Reasoning

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Comparing cardinalities

If ∃ A → B surjection then |A| ≥ |B| If ∃ A → B injection then |A| ≤ |B| If ∃ A → B bijection then |A| = |B|

  • Example. A, B ⊆ Z

Sumset A + B = {a + b | a ∈ A, b ∈ B} Proposition (little theorem) |A + B| ≤ |A| · |B| Proof via surjection We know: |A × B| = |A| · |B|. Goal: Find surjection f : A × B → A + B. So for (a, b) ∈ A × B, need to define f(a, b) CHAT! f(a, b) := a + b Why is this function surjective? The definition of A + B says: A + B = range(f). (Check!)

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B|

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called?

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision.

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision. Notation: For n ≥ 0 we write [n] = {1, . . . , n} Examples: [3] = {1, 2, 3}, [1] = {1}, [0] =

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision. Notation: For n ≥ 0 we write [n] = {1, . . . , n} Examples: [3] = {1, 2, 3}, [1] = {1}, [0] = ∅

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision. Notation: For n ≥ 0 we write [n] = {1, . . . , n} Examples: [3] = {1, 2, 3}, [1] = {1}, [0] = ∅ Bonus Let n ≥ 1. Let S ⊆ [2n] such that |S| ≥ n + 1. Prove: (∃a, b ∈ S)((a b) ∧ (a | b)).

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision. Notation: For n ≥ 0 we write [n] = {1, . . . , n} Examples: [3] = {1, 2, 3}, [1] = {1}, [0] = ∅ Bonus Let n ≥ 1. Let S ⊆ [2n] such that |S| ≥ n + 1. Prove: (∃a, b ∈ S)((a b) ∧ (a | b)). Why |S| ≥ n + 1? Because it is false for |S| = n. What does it mean that it is false?

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision. Notation: For n ≥ 0 we write [n] = {1, . . . , n} Examples: [3] = {1, 2, 3}, [1] = {1}, [0] = ∅ Bonus Let n ≥ 1. Let S ⊆ [2n] such that |S| ≥ n + 1. Prove: (∃a, b ∈ S)((a b) ∧ (a | b)). Why |S| ≥ n + 1? Because it is false for |S| = n. What does it mean that it is false? HW (∀n ≥ 1) find T ⊂ [2n] such that |T| = n and (∀a, b ∈ T)((a | b) ⇒

CMSC-37115 Mathematical Reasoning

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Comparison via injection

If ∃ A → B injection then |A| ≤ |B| This is a famous “principle.” What is it called? Pigeon Hole Principle. If |A| > |B| then every function f : A → B has a collision. Notation: For n ≥ 0 we write [n] = {1, . . . , n} Examples: [3] = {1, 2, 3}, [1] = {1}, [0] = ∅ Bonus Let n ≥ 1. Let S ⊆ [2n] such that |S| ≥ n + 1. Prove: (∃a, b ∈ S)((a b) ∧ (a | b)). Why |S| ≥ n + 1? Because it is false for |S| = n. What does it mean that it is false? HW (∀n ≥ 1) find T ⊂ [2n] such that |T| = n and (∀a, b ∈ T)((a | b) ⇒ (a = b))

CMSC-37115 Mathematical Reasoning

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Counting injections, surjections # functions [k] → [n]

CMSC-37115 Mathematical Reasoning

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Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n]

CMSC-37115 Mathematical Reasoning

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Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

CMSC-37115 Mathematical Reasoning

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SLIDE 37

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections?

CMSC-37115 Mathematical Reasoning

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SLIDE 38

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n]

CMSC-37115 Mathematical Reasoning

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SLIDE 39

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n] What if k < n ?

CMSC-37115 Mathematical Reasoning

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SLIDE 40

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n] What if k < n ? Zero

CMSC-37115 Mathematical Reasoning

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SLIDE 41

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n] What if k < n ? Zero Let k = n. # surjections

CMSC-37115 Mathematical Reasoning

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Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n] What if k < n ? Zero Let k = n. # surjections = # bijections

CMSC-37115 Mathematical Reasoning

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SLIDE 43

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n] What if k < n ? Zero Let k = n. # surjections = # bijections = n!

CMSC-37115 Mathematical Reasoning

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SLIDE 44

Counting injections, surjections # functions [k] → [n] nk # injections [k] → [n] n(n − 1) . . . (n − k + 1) =

n! (n−k)!

Is this true when k > n? What is then the # of injections? # surjections [k] → [n] What if k < n ? Zero Let k = n. # surjections = # bijections = n! Bonus Count the surjections [n + 1] → [n]. HW Count the surjections [k] → [2].

CMSC-37115 Mathematical Reasoning

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Birthday paradox k random people gather for a videoconference. What is the chance that two of them have the same birthday?

CMSC-37115 Mathematical Reasoning

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Birthday paradox k random people gather for a videoconference. What is the chance that two of them have the same birthday? If k = 367 then

CMSC-37115 Mathematical Reasoning

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Birthday paradox k random people gather for a videoconference. What is the chance that two of them have the same birthday? If k = 367 then 100% because

CMSC-37115 Mathematical Reasoning

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Birthday paradox k random people gather for a videoconference. What is the chance that two of them have the same birthday? If k = 367 then 100% because PHP

CMSC-37115 Mathematical Reasoning

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Birthday paradox k random people gather for a videoconference. What is the chance that two of them have the same birthday? If k = 367 then 100% because PHP What is the smallest k s.t. bday match has ≥ 50% chance? Guess a number.

CMSC-37115 Mathematical Reasoning

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Birthday paradox k random people gather for a videoconference. What is the chance that two of them have the same birthday? If k = 367 then 100% because PHP What is the smallest k s.t. bday match has ≥ 50% chance? Guess a number. 23

CMSC-37115 Mathematical Reasoning

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Birthday paradox Model: random function f : [k] → [365] Chance of collision? Let p(k, n) = the probability that a random function f : [k] → [n] is injective. Question: Min k such that p(k, 365) ≤ 1/2

CMSC-37115 Mathematical Reasoning

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Birthday paradox Model: random function f : [k] → [365] Chance of collision? Let p(k, n) = the probability that a random function f : [k] → [n] is injective. Question: Min k such that p(k, 365) ≤ 1/2 Theorem p(23, 365) ≤ 1/2.

CMSC-37115 Mathematical Reasoning

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Birthday paradox Model: random function f : [k] → [365] Chance of collision? Let p(k, n) = the probability that a random function f : [k] → [n] is injective. Question: Min k such that p(k, 365) ≤ 1/2 Theorem p(23, 365) ≤ 1/2. 23 people suffice for a ≥ 50% chance of birthday match.

CMSC-37115 Mathematical Reasoning

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Birthday paradox

We perform an experiment (like, gather a videoconference). Naive probability of an event: number of good outcomes total number of possible outcomes

CMSC-37115 Mathematical Reasoning

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Birthday paradox

We perform an experiment (like, gather a videoconference). Naive probability of an event: number of good outcomes total number of possible outcomes p(k, n): probability that random [k] → [n] function injective p(n, k) = n(n − 1) · · · (n − k + 1) nk

CMSC-37115 Mathematical Reasoning

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SLIDE 56

Birthday paradox

We perform an experiment (like, gather a videoconference). Naive probability of an event: number of good outcomes total number of possible outcomes p(k, n): probability that random [k] → [n] function injective p(n, k) = n(n − 1) · · · (n − k + 1) nk Lemma (∀x ∈ R)(1 + x ≤ ex)

CMSC-37115 Mathematical Reasoning

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SLIDE 57

Birthday paradox

We perform an experiment (like, gather a videoconference). Naive probability of an event: number of good outcomes total number of possible outcomes p(k, n): probability that random [k] → [n] function injective p(n, k) = n(n − 1) · · · (n − k + 1) nk Lemma (∀x ∈ R)(1 + x ≤ ex) Proof: a little calculus.

CMSC-37115 Mathematical Reasoning

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SLIDE 58

Birthday paradox

p(n, k) = n(n − 1)(n − 2)(n − 3) · · · (n − k + 1) nk

CMSC-37115 Mathematical Reasoning

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SLIDE 59

Birthday paradox

p(n, k) = n(n − 1)(n − 2)(n − 3) · · · (n − k + 1) nk = n n · n − 1 n · n − 2 n · n − 3 n · · · n − k + 1 n

CMSC-37115 Mathematical Reasoning

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SLIDE 60

Birthday paradox

p(n, k) = n(n − 1)(n − 2)(n − 3) · · · (n − k + 1) nk = n n · n − 1 n · n − 2 n · n − 3 n · · · n − k + 1 n =

  • 1 − 1

n

  • ·
  • 1 − 2

n

  • ·
  • 1 − 3

n

  • · · ·
  • 1 − n − k + 1

n

  • CMSC-37115

Mathematical Reasoning

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SLIDE 61

Birthday paradox

p(n, k) = n(n − 1)(n − 2)(n − 3) · · · (n − k + 1) nk = n n · n − 1 n · n − 2 n · n − 3 n · · · n − k + 1 n =

  • 1 − 1

n

  • ·
  • 1 − 2

n

  • ·
  • 1 − 3

n

  • · · ·
  • 1 − n − k + 1

n

  • ≤

e−1/n · e−2/n · e−3/n · · · e−(k−1)/n

CMSC-37115 Mathematical Reasoning

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SLIDE 62

Birthday paradox

p(n, k) = n(n − 1)(n − 2)(n − 3) · · · (n − k + 1) nk = n n · n − 1 n · n − 2 n · n − 3 n · · · n − k + 1 n =

  • 1 − 1

n

  • ·
  • 1 − 2

n

  • ·
  • 1 − 3

n

  • · · ·
  • 1 − n − k + 1

n

  • ≤

e−1/n · e−2/n · e−3/n · · · e−(k−1)/n = e−(1+2+3+···+(k−1))/n = e−k(k−1)/(2n)

CMSC-37115 Mathematical Reasoning

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SLIDE 63

Birthday paradox

p(n, k) = n(n − 1)(n − 2)(n − 3) · · · (n − k + 1) nk = n n · n − 1 n · n − 2 n · n − 3 n · · · n − k + 1 n =

  • 1 − 1

n

  • ·
  • 1 − 2

n

  • ·
  • 1 − 3

n

  • · · ·
  • 1 − n − k + 1

n

  • ≤

e−1/n · e−2/n · e−3/n · · · e−(k−1)/n = e−(1+2+3+···+(k−1))/n = e−k(k−1)/(2n) Theorem The probability that a random [k] → [n] function is an injection: p(k, n) ≤ e−k(k−1)/(2n)

CMSC-37115 Mathematical Reasoning

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SLIDE 64

Birthday paradox Theorem The probability that a random [k] → [n] function is an injection: p(k, n) ≤ e−k(k−1)/(2n) Corollary p(23, 365) ≤ e−23·22/730 = e−506/730 < 0.4999982

CMSC-37115 Mathematical Reasoning

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SLIDE 65

Birthday paradox Theorem The probability that a random [k] → [n] function is an injection: p(k, n) ≤ e−k(k−1)/(2n) Corollary p(23, 365) ≤ e−23·22/730 = e−506/730 < 0.4999982 ∴ the probability that 23 people will have a birthday match, is at least 1 − 0.4999982 > 0.500001

CMSC-37115 Mathematical Reasoning