CSci 2011 Discrete Mathematics Lecture 9, 10

Recap Propositional operation summary Check translation Definition Tautology, Contradiction, logical equicalence not and or conditional Bi-conditional p q p q pq pq pq pq T F CSci 2011

Recap CSci 2011 p  T  p p  F  p (p  q)  r  p  (q  r) Identity Laws (p  q)  r  p  (q  r) (p  q)  r  p  (q  r) Associative laws p  T  T p  F  F Domination Law p  (q  r)  (p  q)  (p  r) p  (q  r)  (p  q)  (p  r) Distributive laws p  p  p p  p  p Idempotent Laws  (p  q)   p   q  (p  q)   p   q De Morgan’s laws ( p)  p Double negation law p  (p  q)  p p  (p  q)  p Absorption laws p  q  q  p p  q  q  p Commutative Laws p   p  T p   p  F Negation lows pq  pq Definition of Implication p  q  (p  q)  (q  p) Definition of Biconditional CSci 2011

Recap Quantifiers CSci 2011 Nested quantifiers Universal quantifier: x P(x) Negating quantifiers ¬x P(x) = x ¬P(x) ¬x P(x) = x ¬P(x) xy P(x, y) Nested quantifiers xy P(x, y): “For all x, there exists a y such that P(x,y)” xy P(x,y): There exists an x such that for all y P(x,y) is true” ¬x P(x) = x ¬P(x), ¬x P(x) = x ¬P(x) CSci 2011

Recap p p  q  q  q   p q  r  p  r p  q  p  p  q p  q  p Modus ponens p p  q  q Modus tollens  q   p Hypothetical syllogism q  r  p  r Disjunctive syllogism p  q  p Addition  p  q Simplification p  q  p Conjunction q  p  q Resolution  p  r  q  r CSci 2011

Recap CSci 2011 p→q Proof by contradiction Direct Proof: Assume p is true. Show that q is also true. Indirect Proof: Assume ¬q is true. Show that p is true. Proof by contradiction Proving p: Assume p is not true. Find a contradiction. Proving p→q ¬(p→q)   (p  q)  p  ¬q Assume p is tue and q is not true. Find a contradiction. Proof by Cases: [(p1p2…pn)q]  [(p1q)(p2q)…(pnq)] If and only if proof: pq (p→q)(q→p) Existence Proof: Constructive vs. Non-constructive Proof Uniqueness: 1) Show existence 2) find contradiction if not unique Forward and backward reasoning Counterexample CSci 2011

What is a set? CSci 2011 A set is a unordered collection of “objects” Order does not matter Sets do not have duplicate elements set-builder notation: D = {x | x is prime and x > 2} a  S if an element a is an element of a set S. a  S if not. U is the universal set. empty (or null) set  = { }, if a set has zero elements. Sets can contain other sets S = {{1},{2},{3}} V = {{{1},{{2}}},{{{3}}},{{1},{{2}},{{{3}}}}} Venn Diagram S  T if  x (x  S  x  T) S = T if S  T and if S  T. S S  S,   S CSci 2011

Set Equality, Subsets Two sets are equal if they have the same elements {1, 2, 3, 4, 5} = {5, 4, 3, 2, 1} {1, 2, 3, 2, 4, 3, 2, 1} = {4, 3, 2, 1} Two sets are not equal if they do not have the same elements {1, 2, 3, 4, 5} ≠ {1, 2, 3, 4} If all the elements of a set S are also elements of a set T, then S is a subset of T If S = {2, 4, 6}, T = {1, 2, 3, 4, 5, 6, 7}, S is a subset of T This is specified by S  T meaning that  x (x  S  x  T) For any set S, S  S (S S  S) For any set S,   S (S   S) CSci 2011

Proper Subsets If S is a subset of T, and S is not equal to T, then S is a proper subset of T Can be written as: R  T and R  T Let T = {0, 1, 2, 3, 4, 5} If S = {1, 2, 3}, S is not equal to T, and S is a subset of T A proper subset is written as S  T Let Q = {4, 5, 6}. Q is neither a subset or T nor a proper subset of T The difference between “subset” and “proper subset” is like the difference between “less than or equal to” and “less than” for numbers CSci 2011

Set cardinality The cardinality of a set is the number of elements in a set, written as |A| Examples Let R = {1, 2, 3, 4, 5}. Then |R| = 5 || = 0 Let S = {, {a}, {b}, {a, b}}. Then |S| = 4 CSci 2011

Power Sets Given S = {0, 1}. All the possible subsets of S?  (as it is a subset of all sets), {0}, {1}, and {0, 1} The power set of S (written as P(S)) is the set of all the subsets of S P(S) = { , {0}, {1}, {0,1} } Note that |S| = 2 and |P(S)| = 4 Let T = {0, 1, 2}. The P(T) = { , {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2} } Note that |T| = 3 and |P(T)| = 8 P() = {  } Note that || = 0 and |P()| = 1 If a set has n elements, then the power set will have 2n elements CSci 2011

Tuples In 2-dimensional space, it is a (x, y) pair of numbers to specify a location In 3-dimensional (1,2,3) is not the same as (3,2,1) – space, it is a (x, y, z) triple of numbers In n-dimensional space, it is a n-tuple of numbers Two-dimensional space uses pairs, or 2-tuples Three-dimensional space uses triples, or 3-tuples Note that these tuples are ordered, unlike sets the x value has to come first +x +y (2,3) CSci 2011

Cartesian products A Cartesian product is a set of all ordered 2-tuples where each “part” is from a given set Denoted by A x B, and uses parenthesis (not curly brackets) For example, 2-D Cartesian coordinates are the set of all ordered pairs Z x Z Recall Z is the set of all integers This is all the possible coordinates in 2-D space Example: Given A = { a, b } and B = { 0, 1 }, what is their Cartiesian product? C = A x B = { (a,0), (a,1), (b,0), (b,1) } Formal definition of a Cartesian product: A x B = { (a,b) | a  A and b  B } CSci 2011

Cartesian Products 2 All the possible grades in this class will be a Cartesian product of the set S of all the students in this class and the set G of all possible grades Let S = { Alice, Bob, Chris } and G = { A, B, C } D = { (Alice, A), (Alice, B), (Alice, C), (Bob, A), (Bob, B), (Bob, C), (Chris, A), (Chris, B), (Chris, C) } The final grades will be a subset of this: { (Alice, C), (Bob, B), (Chris, A) } Such a subset of a Cartesian product is called a relation (more on this later in the course) CSci 2011

ch2.2 Set Operations CSci 2011

Set operations: Union Formal definition for the union of two sets: A U B = { x | x  A or x  B } Further examples {1, 2, 3} U {3, 4, 5} = {1, 2, 3, 4, 5} {a, b} U {3, 4} = {a, b, 3, 4} {1, 2} U  = {1, 2} Properties of the union operation A U  = A Identity law A U U = U Domination law A U A = A Idempotent law A U B = B U A Commutative law A U (B U C) = (A U B) U C Associative law CSci 2011

Set operations: Intersection Formal definition for the intersection of two sets: A ∩ B = { x | x  A and x  B } Examples {1, 2, 3} ∩ {3, 4, 5} = {3} {a, b} ∩ {3, 4} =  {1, 2} ∩  =  Properties of the intersection operation A ∩ U = A Identity law A ∩  =  Domination law A ∩ A = A Idempotent law A ∩ B = B ∩ A Commutative law A ∩ (B ∩ C) = (A ∩ B) ∩ C Associative law CSci 2011

Disjoint sets Formal definition for disjoint sets: two sets are disjoint if their intersection is the empty set Further examples {1, 2, 3} and {3, 4, 5} are not disjoint {a, b} and {3, 4} are disjoint {1, 2} and  are disjoint Their intersection is the empty set  and  are disjoint! CSci 2011

Set operations: Difference Formal definition for the difference of two sets: A - B = { x | x  A and x  B } Further examples {1, 2, 3} - {3, 4, 5} = {1, 2} {a, b} - {3, 4} = {a, b} {1, 2} -  = {1, 2} The difference of any set S with the empty set will be the set S CSci 2011

Complement sets Formal definition for the complement of a set: A = { x | x  A } = Ac Or U – A, where U is the universal set Further examples (assuming U = Z) {1, 2, 3}c = { …, -2, -1, 0, 4, 5, 6, … } Properties of complement sets (Ac)c = A Complementation law A U Ac = U Complement law A ∩ Ac =  Complement law CSci 2011

Set identities CSci 2011 A = A AU = A Identity Law AU = U A =  Domination law AA = A AA = A Idempotent Law (Ac)c = A Complement Law AB = BA AB = BA Commutative Law (AB)c = AcBc (AB)c = AcBc De Morgan’s Law A(BC) = (AB)C A(BC) = (AB)C Associative Law A(BC) = (AB)(AC) A(BC) = (AB)(AC) Distributive Law A(AB) = A A(AB) = A Absorption Law A  Ac = U A  Ac =  CSci 2011

How to prove a set identity For example: A∩B=B-(B-A) Four methods: Use the basic set identities Use membership tables Prove each set is a subset of each other Use set builder notation and logical equivalences CSci 2011

What we are going to prove… A∩B=B-(B-A) A B B-(B-A) A∩B B-A CSci 2011

Proof by Set Identities A  B = A - (A - B) Proof) A - (A - B) = A - (A  Bc) = A  (A  Bc)c = A  (Ac  B) = (A  Ac)  (A  B) =   (A  B) = A  B CSci 2011

Showing each is a subset of the others (A  B)c = Ac  Bc Proof) Want to prove that (A  B)c  Ac  Bc and Ac  Bc  (A  B)c x  (A  B)c  x  (A  B)   (x  A  B)   (x  A  x  B)   (x  A)   (x  B)  x  A  x  B  x  Ac  x  Bc  x  Ac  Bc CSci 2011

Examples Let A, B, and C be sets. Show that: (AUB)  (AUBUC) (A∩B∩C)  (A∩B) (A-B)-C  A-C (A-C) ∩ (C-B) =  CSci 2011

ch2.3 Functions CSci 2011

Definition of a function A function takes an element from a set and maps it to a UNIQUE element in another set f maps R to Z R Z Domain Co-domain f f(4.3) 4.3 4 Pre-image of 4 Image of 4.3 CSci 2011

A string length function More functions The image of “a” A pre-image of 1 Domain Co-domain A B C D F Alice Bob Chris Dave Emma A class grade function 1 2 3 4 5 “a” “bb“ “cccc” “dd” “e” A string length function CSci 2011

Also not a valid function! Even more functions Range 1 2 3 4 5 a e i o u Some function… 1 2 3 4 5 “a” “bb“ “cccc” “dd” “e” Not a valid function! Also not a valid function! CSci 2011

Function arithmetic Let f1(x) = 2x Let f2(x) = x2 f1+f2 = (f1+f2)(x) = f1(x)+f2(x) = 2x+x2 f1*f2 = (f1*f2)(x) = f1(x)*f2(x) = 2x*x2 = 2x3 CSci 2011

One-to-one functions A function is one-to-one if each element in the co-domain has a unique pre-image Formal definition: A function f is one-to-one if f(x) = f(y) implies x = y. 1 2 3 4 5 a e i o A one-to-one function 1 2 3 4 5 a e i o A function that is not one-to-one CSci 2011

More on one-to-one Injective is synonymous with one-to-one “A function is injective” A function is an injection if it is one-to-one Note that there can be un-used elements in the co-domain 1 2 3 4 5 a e i o A one-to-one function CSci 2011

A function that is not onto Onto functions A function is onto if each element in the co-domain is an image of some pre-image Formal definition: A function f is onto if for all y  C, there exists x  D such that f(x)=y. 1 2 3 4 a e i o u An onto function 1 2 3 4 5 a e i o A function that is not onto CSci 2011

More on onto Surjective is synonymous with onto “A function is surjective” A function is an surjection if it is onto Note that there can be multiply used elements in the co-domain 1 2 3 4 a e i o u An onto function CSci 2011

Onto vs. one-to-one Are the following functions onto, one-to-one, both, or neither? 1 2 3 4 a b c 1 2 3 4 a b c d 1 2 3 4 a b c 1-to-1, not onto Both 1-to-1 and onto Not a valid function 1 2 3 a b c d 1 2 3 4 a b c d Onto, not 1-to-1 Neither 1-to-1 nor onto CSci 2011

Bijections Consider a function that is both one-to-one and onto: Such a function is a one-to-one correspondence, or a bijection 1 2 3 4 a b c d CSci 2011

Identity functions A function such that the image and the pre-image are ALWAYS equal f(x) = 1*x f(x) = x + 0 The domain and the co-domain must be the same set CSci 2011

Inverse functions Let f(x) = 2*x R f R f-1 f(4.3) 4.3 8.6 f-1(8.6) Then f-1(x) = x/2 CSci 2011

More on inverse functions Can we define the inverse of the following functions? An inverse function can ONLY be done defined on a bijection 1 2 3 4 a b c 1 2 3 a b c d What is f-1(2)? Not onto! What is f-1(2)? Not 1-to-1! CSci 2011

Compositions of functions (f ○ g)(x) = f(g(x)) f ○ g A B C g f g(a) f(b) a f(g(a)) b = g(a) (f ○ g)(a) CSci 2011

Compositions of functions Let f(x) = 2x+3 Let g(x) = 3x+2 f ○ g R R R g f g(1) f(5) f(g(1))=13 1 g(1)=5 (f ○ g)(1) f(g(x)) = 2(3x+2)+3 = 6x+7 CSci 2011

Compositions of functions Does f(g(x)) = g(f(x))? Let f(x) = 2x+3 Let g(x) = 3x+2 f(g(x)) = 2(3x+2)+3 = 6x+7 g(f(x)) = 3(2x+3)+2 = 6x+11 Function composition is not commutative! Not equal! CSci 2011

Useful functions Floor: x means take the greatest integer less than or equal to the number Ceiling: x means take the lowest integer greater than or equal to the number round(x) =  x+0.5  CSci 2011

Ceiling and floor properties Let n be an integer (1a) x = n if and only if n ≤ x < n+1 (1b) x = n if and only if n-1 < x ≤ n (1c) x = n if and only if x-1 < n ≤ x (1d) x = n if and only if x ≤ n < x+1 (2) x-1 < x ≤ x ≤ x < x+1 (3a) -x = - x (3b) -x = - x (4a) x+n = x+n (4b) x+n = x+n CSci 2011

Ceiling property proof Prove rule 4a: x+n = x+n Where n is an integer Will use rule 1a: x = n if and only if n ≤ x < n+1 Direct proof! Let m = x Thus, m ≤ x < m+1 (by rule 1a) Add n to both sides: m+n ≤ x+n < m+n+1 By rule 1a, m+n = x+n Since m = x, m+n also equals x+n Thus, x+n = m+n = x+n CSci 2011

Factorial Factorial is denoted by n! n! = n * (n-1) * (n-2) * … * 2 * 1 Thus, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 Note that 0! is defined to equal 1 CSci 2011

Proving Function problems Let f be an invertible function from Y to Z Let g be an invertible function from X to Y Show that the inverse of f○g is: (f○g)-1 = g-1 ○ f-1 (Pf) Thus, we want to show, for all zZ and xX ((f  g)  (g-1  f-1)) (x) = x and ((f-1  g-1)  (g  f)) (z) = z ((f  g)  (g-1  f-1)) (x) = (f  g) (g-1  f-1)) (x)) = (f  g) g-1 f-1(x))) = (f g g-1 f-1(x))))) = (f f-1(x)) = x The second equality is similar CSci 2011