Any Wizard of Oz fans? It is always best to start at the beginning - PDF document

Any Wizard of Oz fans? “It is always best to start at the beginning” Discrete Math Basics -- Glynda, the good witch of the North Outline Sets • Sets • A set is a collection of elements. • Relations – Finite set – has finite number of elements • A = { 1, 3, 5, 6, 9 } • Proofs – Infinite Set – has infinite number of elements • B = { x | x is an odd integer } – Null (Empty) set – has no elements. • ∅ Sets Set Operations • Membership • Union: – A ∪ B = set consisting of all elements in either A or B – If an element x is a member of a set A, we write: or both. • x ∈ A. – If an element x is not a member of a set A, we write: • Intersection • x ∉ A – A ∩ B = set of elements that are in both A and B • Subsets: – If A ∩ B = ∅ , A and B are disjoint. – A is a subset of B if all elements of A are also in B. • Difference – A ⊆ B if x ∈ A then x ∈ B – A – B = set of all elements of A that are not elements of – A = B is the same as saying A ⊆ B and B ⊆ A. B • Power Set • Complement (wrt a universal set U) – 2 A = Set of all subsets of A – A’ = All elements in U that are not in A – U - A

Set Operations Cartesian Product of Sets • A x B = set of ordered pairs (a,b) such that • A = { 1, 3, 5, 6, 9 } – a ∈ A and b ∈ B. • B = { x | x is an odd integer } • A x B x C = set of ordered triplets (a, b, c) such that • A ∪ B = { x | x = 6 or x is an odd integer} – a ∈ A and b ∈ B and c ∈ C • A ∩ B = { 1, 3, 5, 9 } • In general: A 1 x A 2 x … x A n = the set of all • A – B = { 6 } n-tuples (a 1 , a 2 , …, a n ) such that • B’ = { x | x is an even integer } – a i ∈ A i for all i. – With respect to the Universal set of Positive integers. Relations on sets Functions • Means to relate or associate a member of • Functions are restricted relations one set with a member of another: – f: A → B – R : A → B – Relation is simply a subset of A x B – Every element of A is associated with exactly 1 • If a ∈ A is related to b ∈ B then (a, b) ∈ R element of B • Can also be written aRb • Or R(a) contains b Functions Functions • A function, f: A → B is: • f : A → B – Onto (Surjective) if every element of B is – A and B can themselves be Cartesian Products related to at least one element of A – f: A x B → C – One-to-one (Injective) if every element of B is • Elements of f are ((a,b), c) where a ∈ A, b ∈ B, c ∈ related to at most one element of A. C • f (a,b) = c – A bijection if it is both one-to-one and onto • Every element in B is related to one and only one element of A.

Equivalence Relations Equivalence Relations • An equivalence relation on A (R: A → A) • If R is a relation on A (R: A → A), R is an partitions the elements of A into disjoint equivalence relation if: equivalence classes. – R is reflexive – (a, a) ∈ R for all a in A – For each class, elements in the class are related only to – R is symmetric – if (a,b) ∈ R then (b,a) ∈ R all elements in the same partition. – Example: – R is transitive – if (a, b) ∈ R and (b,c) ∈ R then • ≡ 4 : Ν → Ν : x ≡ 4 y if ( x mod 4) == (y mod 4 ) (a,c) ∈ R • [0] = { 0, 4, 8, 12, …} • [1] = { 1, 5, 9, 13, …} • [2] = { 2, 7, 10, 15, …} • [3] = { 3, 8, 11, 16, …} Functions and Relations Mathematical Proofs • Questions? • An argument that something is true – If X, then Y • Types of proofs • Next: proofs – Direct/Constructive proofs – Proof by contradiction – Proof by induction Direct Proofs Direct Proofs • If X, then Y • Example: – For integers a,b : If a and b are odd, then ab is • Assume X is true, show directly that Y is odd. true. – Given: a and b are odd integers • There exists integer x such that a= 2x +1 • There exists integer y such that b = 2y + 1 – Must prove: a times b is also odd • There exists integer z such that ab = 2z + 1

Direct Proofs Proof by contradiction • Perform the multiplication directly • If X, then Y – ab = (2x + 1)(2y + 1) • With proof by contraction = 4xy + 2x + 2y + 1 – You assume that Y is false = 2(2xy + x + y) + 1 – Then derive a contradiction to a fact known to So z = 2xy + x + y be true – If we find a contradiction, then we know our Not only did you prove that a z exists, you constructed an initial assumption (I.e. Y is false) must be “algorithm” for generating this z. invalid and thus Y must be true. This is an example of a constructive proof. Proof by contradiction Proof by contradiction • If X, then Y -- Assume Y is false • Example 1. Assume that A ∩ C ≠ ∅ – A, B, and C are sets. If A ∩ B = ∅ and C ⊆ • Meaning, there is an element x which is in both A and C B, then A ∩ C = ∅ 2. Since C is a subset of B then x, being a member of C must a member of B as well. – Given: 3. However, by #1, x ∈ A. • A and B have nothing in common 4. x ∈ A and x ∈ B implies that A ∩ B ≠ ∅ • C is a subset of B 5. This contradicts X – Must show: 6. The assumption in #1 must be invalid and as such A ∩ • A and C have nothing in common C = ∅ Proof by Induction Proof by Induction • Used to prove statements involving an • Notice there are two elements to the integer that we wish to prove true for all problem to be proven: integers greater than a given integer – Statement involving an integer, P(n) – Example: – A specific integer n 0 for which we are trying to + n n ( n 1 ) ∑ show that P(n) is true for all values greater than = + + + + + = i 1 2 3 4 n L 2 or equal to. i = 1 For all n ≥ 0

Proof by induction Proof by induction • How : • Steps to an inductive proof: – Let P(n) is some statement involving an 1. Basis step: integer, n. To prove that P(n) is true for Show P(n) is true when n=n 0 every integer ≥ a given integer n 0 , it is 2. Induction hypothesis sufficient to show these two things: Assume that P(n) is true for some k ≥ n 0 1. P(n 0 ) is true 3. Inductive step 2. For any k ≥ n 0 , if P(k) is true, then P(k+1) is true. Prove P(n) is true for n = k+1 using the induction hypothesis. Proof by induction Proof by Induction 2. Inductive Hypothesis: Assume true for n=k • Example: + k k ( k 1 ) – Show that for n ≥ 0 ∑ = i + n n ( n 1 ) ∑ 2 = + + + + + = i 1 2 3 4 L n i = 1 2 = i 1 3. Inductive Step: Show true for n=k +1 1. Basis step: Show true for n = 0 + + + k 1 ( k 1 )( k 2 ) ∑ = i 0 + ∑ 0 ( 0 1 ) 2 = i 0 = = i 1 0 2 = i 1 Proof by induction Proof by induction k + 1 k ∑ = ∑ + + i i ( k 1 ) • Steps to an inductive proof: = = i 1 i 1 1. Basis step: By the induction hypothesis Show P(n) is true when n=n 0 + k k ( k 1 ) ∑ so… = i 2. Induction hypothesis 2 = i 1 Assume that P(n) is true for some k ≥ n 0 + + + + k + 1 k ( k 1 ) k ( k 1 ) 2 ( k 1 ) ∑ = + + = i ( k 1 ) 3. Inductive step 2 2 = i 1 Prove P(n) is true for n = k+1 using the induction + + + + k 2 2 k 2 ( k 1 )( k 2 ) hypothesis. = = 2 2

Summary Reminder • Proof • Next class… – Direct / constructive – Quiz on all this discrete math. – Proof by contradiction – Start to look at strings and languages – Proof by induction • Any questions?