Intro to Mathematical Reasoning via Discrete Mathematics - - PowerPoint PPT Presentation

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intro to mathematical reasoning via discrete mathematics

Intro to Mathematical Reasoning via Discrete Mathematics - - PowerPoint PPT Presentation

Jun 24, 2023 245 likes •626 views

Intro to Mathematical Reasoning via Discrete Mathematics CMSC-37115 Instructor: Laszlo Babai University of Chicago Week 1, Thursday, October 1, 2020 CMSC-37115 Mathematical Reasoning Sets, subsets setmaker bar such that A = { x 2 | 0

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Intro to Mathematical Reasoning via Discrete Mathematics

CMSC-37115 Instructor: Laszlo Babai University of Chicago Week 1, Thursday, October 1, 2020

CMSC-37115 Mathematical Reasoning

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Sets, subsets setmaker bar “such that” A = {x2 | 0 ≤ x ≤ 3} = {0, 1, 4, 9}

CMSC-37115 Mathematical Reasoning

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Sets, subsets setmaker bar “such that” A = {x2 | 0 ≤ x ≤ 3} = {0, 1, 4, 9} A = {a, b, c, d, e} B = {a, b, e} =⇒ B ⊆ A subset B ⊆ A if (∀x)((x ∈ B) =⇒ (x ∈ A))

CMSC-37115 Mathematical Reasoning

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Sets, subsets setmaker bar “such that” A = {x2 | 0 ≤ x ≤ 3} = {0, 1, 4, 9} A = {a, b, c, d, e} B = {a, b, e} =⇒ B ⊆ A subset B ⊆ A if (∀x)((x ∈ B) =⇒ (x ∈ A)) DO The “subset” relation is

CMSC-37115 Mathematical Reasoning

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Sets, subsets setmaker bar “such that” A = {x2 | 0 ≤ x ≤ 3} = {0, 1, 4, 9} A = {a, b, c, d, e} B = {a, b, e} =⇒ B ⊆ A subset B ⊆ A if (∀x)((x ∈ B) =⇒ (x ∈ A)) DO The “subset” relation is transitive A ⊆ B ⊆ C =⇒ A ⊆ C

CMSC-37115 Mathematical Reasoning

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Sets, subsets setmaker bar “such that” A = {x2 | 0 ≤ x ≤ 3} = {0, 1, 4, 9} A = {a, b, c, d, e} B = {a, b, e} =⇒ B ⊆ A subset B ⊆ A if (∀x)((x ∈ B) =⇒ (x ∈ A)) DO The “subset” relation is transitive A ⊆ B ⊆ C =⇒ A ⊆ C A = B if A ⊆ B and B ⊆ A, i.e., A = B if (∀x)(x ∈ A ⇔ x ∈ B) Example: {a, c, a, b, c} = {a, b, c}.

CMSC-37115 Mathematical Reasoning

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Sets, subsets setmaker bar “such that” A = {x2 | 0 ≤ x ≤ 3} = {0, 1, 4, 9} A = {a, b, c, d, e} B = {a, b, e} =⇒ B ⊆ A subset B ⊆ A if (∀x)((x ∈ B) =⇒ (x ∈ A)) DO The “subset” relation is transitive A ⊆ B ⊆ C =⇒ A ⊆ C A = B if A ⊆ B and B ⊆ A, i.e., A = B if (∀x)(x ∈ A ⇔ x ∈ B) Example: {a, c, a, b, c} = {a, b, c}. Powerset of A: P(A) = {all subsets of A}

CMSC-37115 Mathematical Reasoning

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Operations with sets DEFINITIONS Union A ∪ B = {x | (x ∈ A) ∨ (x ∈ B)} Intersection A ∩ B = {x | (x ∈ A) ∧ (x ∈ B)} Difference A \ B = {x | (x ∈ A) ∧ (x B)} A, B ⊆ Ω “universe” Complement A = Ω \ A

CMSC-37115 Mathematical Reasoning

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Operations with sets DEFINITIONS Union A ∪ B = {x | (x ∈ A) ∨ (x ∈ B)} Intersection A ∩ B = {x | (x ∈ A) ∧ (x ∈ B)} Difference A \ B = {x | (x ∈ A) ∧ (x B)} A, B ⊆ Ω “universe” Complement A = Ω \ A

CMSC-37115 Mathematical Reasoning

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Indentities for set operations DO: Distributivity A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

CMSC-37115 Mathematical Reasoning

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Indentities for set operations DO: Distributivity A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) De Morgan’s laws A ∪ B = A ∩ B A ∩ B = A ∪ B

CMSC-37115 Mathematical Reasoning

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Indentities for set operations DO: Distributivity A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) De Morgan’s laws A ∪ B = A ∩ B A ∩ B = A ∪ B |A| cardinality of A (size of A, number of elements of A) Example: |{a, c, a, b, c}| = 3

CMSC-37115 Mathematical Reasoning

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Indentities for cardinalities DO Modular identity |A ∩ B| + |A ∪ B| = |A| + |B|

CMSC-37115 Mathematical Reasoning

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Indentities for cardinalities DO Modular identity |A ∩ B| + |A ∪ B| = |A| + |B| DO |A| = n =⇒ |P(A)| =

CMSC-37115 Mathematical Reasoning

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Indentities for cardinalities DO Modular identity |A ∩ B| + |A ∪ B| = |A| + |B| DO |A| = n =⇒ |P(A)| = 2n

CMSC-37115 Mathematical Reasoning

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Indentities for cardinalities DO Modular identity |A ∩ B| + |A ∪ B| = |A| + |B| DO |A| = n =⇒ |P(A)| = 2n Cartesian product A × B = {(a, b) | a ∈ A, b ∈ B}

CMSC-37115 Mathematical Reasoning

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Indentities for cardinalities DO Modular identity |A ∩ B| + |A ∪ B| = |A| + |B| DO |A| = n =⇒ |P(A)| = 2n Cartesian product A × B = {(a, b) | a ∈ A, b ∈ B} DO |A × B| = |A| · |B|

CMSC-37115 Mathematical Reasoning

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Indentities for cardinalities DO Modular identity |A ∩ B| + |A ∪ B| = |A| + |B| DO |A| = n =⇒ |P(A)| = 2n Cartesian product A × B = {(a, b) | a ∈ A, b ∈ B} DO |A × B| = |A| · |B|

CMSC-37115 Mathematical Reasoning

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Sumsets DEF shifting A ⊆ Z, b ∈ Z A + b = {a + b | a ∈ A} Example: If A = {1, 2, 4} then A + 2 = {3, 4, 6}

CMSC-37115 Mathematical Reasoning

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Sumsets DEF shifting A ⊆ Z, b ∈ Z A + b = {a + b | a ∈ A} Example: If A = {1, 2, 4} then A + 2 = {3, 4, 6} DEF sumset A, B ⊆ Z A + B = {a + b | a ∈ A, b ∈ B} Example B = {2, 4} then A + B = (A + 2) ∪ (A + 4) = {3, 4, 6, 5, 6, 8} = {3, 4, 5, 6, 8}

CMSC-37115 Mathematical Reasoning

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Sumsets DEF sumset A, B ⊆ Z A + B = {a + b | a ∈ A, b ∈ B} True/false |A + B| ≥ |A|

CMSC-37115 Mathematical Reasoning

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Sumsets DEF sumset A, B ⊆ Z A + B = {a + b | a ∈ A, b ∈ B} True/false |A + B| ≥ |A| A + ∅ =

CMSC-37115 Mathematical Reasoning

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Sumsets DEF sumset A, B ⊆ Z A + B = {a + b | a ∈ A, b ∈ B} True/false |A + B| ≥ |A| A + ∅ = ∅

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≤

CMSC-37115 Mathematical Reasoning

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Sumsets DEF sumset A, B ⊆ Z A + B = {a + b | a ∈ A, b ∈ B} True/false |A + B| ≥ |A| A + ∅ = ∅

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≤ 15

CMSC-37115 Mathematical Reasoning

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Sumsets DEF sumset A, B ⊆ Z A + B = {a + b | a ∈ A, b ∈ B} True/false |A + B| ≥ |A| A + ∅ = ∅

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≤ 15

DO |A + B| ≤ |A| · |B|

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ?

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6}

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assuming

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assumingA, B ∅ 2A := {2a | a ∈ A}

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assumingA, B ∅ 2A := {2a | a ∈ A} kA = {ka | a ∈ A} Example: 2Z =

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assumingA, B ∅ 2A := {2a | a ∈ A} kA = {ka | a ∈ A} Example: 2Z = {even numbers}

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assumingA, B ∅ 2A := {2a | a ∈ A} kA = {ka | a ∈ A} Example: 2Z = {even numbers} 2Z + 1 =

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assumingA, B ∅ 2A := {2a | a ∈ A} kA = {ka | a ∈ A} Example: 2Z = {even numbers} 2Z + 1 = {odd numbers}

CMSC-37115 Mathematical Reasoning

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Sumset

Question. |A| = 3, |B| = 5 =⇒ |A + B| ≥ 7

Example when |A + B| = 7 ? A = {0, 1, 2}, B = {0, 1, 2, 3, 4}, A + B = {0, 1, 2, 3, 4, 5, 6} HW (1/12) |A + B| ≥ |A| + |B| − 1 assumingA, B ∅ 2A := {2a | a ∈ A} kA = {ka | a ∈ A} Example: 2Z = {even numbers} 2Z + 1 = {odd numbers} DO Find a small set B s.t. 2Z + B = Z

CMSC-37115 Mathematical Reasoning