Week 15 - Wednesday

 What did we talk about last time?  Review first third of course

 Consider the following shape to the right:  Now, consider the next shape, made up of pieces of exactly the same size:  We have created space out of nowhere!  How is this possible?

 To prove  x  D  P(x) we need to find at least one element of D that makes P(x) true  To disprove  x  D, P(x)  Q(x), we need to find an x that makes P(x) true and Q(x) false

 If the domain is finite, we can use the method of exhaustion, by simply trying every element  Otherwise, we can use a direct proof 1. Express the statement to be proved in the form  x  D, if P(x) then Q(x) 2. Suppose that x is some specific (but arbitrarily chosen) element of D for which P(x) is true 3. Show that the conclusion Q(x) is true by using definitions, other theorems, and the rules for logical inference  Direct proofs should start with the word Proof, end with the word QED, and have a justification next to every step in the argument  For proofs with cases, number each case clearly and show that you have proved the conclusion for all possible cases

 If n is an integer, then:  n is even   k  Z  n = 2k  n is odd   k  Z  n = 2k + 1  If n is an integer where n > 1, then:  n is prime   r  Z +,  s  Z +, if n = r  s, then r = 1 or s = 1  n is composite   r  Z +,  s  Z +  n = r  s and r  1 and s  1  r is rational   a, b  Z  r = a/b and b  0  For n, d  Z,  n is divisible by d   k  Z  n = dk  For any real number x, the floor of x, written  x , is defined as follows:   x  = the unique integer n such that n ≤ x < n + 1  For any real number x, the ceiling of x, written  x , is defined as follows:   x  = the unique integer n such that n – 1 < x ≤ n

 Unique factorization theorem: For any integer n > 1, there exist a positive integer k, distinct prime numbers p 1, p 2, …, p k, and positive integers e 1, e 2, …, e k such that  Quotient remainder theorem: For any integer n and any positive integer d, there exist unique integers q and r such that  n = dq + r and 0 ≤ r < d

 Prove or disprove the following statements:  For all integers a, b, and c, if a | b and a | c then a | (2b − 3c)  For all integers a, b, and c, if a | (b + c) then a | b or a | c  Use the quotient-remainder theorem with d = 3 to prove that the square of any integer has the form 3k or 3k + 1 for some integer k

 Mathematical sequences can be represented in expanded form or with explicit formulas  Examples:  2, 5, 10, 17, 26, …  a i = i 2 + 1, i ≥ 1  Summation notation is used to describe a summation of some part of a series  Product notation is used to describe a product of some part of a series

 To prove a statement of the following form:   n  Z, where n  a, property P(n) is true  Use the following steps: 1. Basis Step: Show that the property is true for P(a) 2. Induction Step: ▪ Suppose that the property is true for some n = k, where k  Z, k  a ▪ Now, show that, with that assumption, the property is also true for k + 1

 Using recursive definitions to generate sequences  Writing a recursive definition based on a sequence  Using mathematical induction to show that a recursive definition and an explicit definition are equivalent

 Expand the recursion repeatedly without combining like terms  Find a pattern in the expansions  When appropriate, employ formulas to simplify the pattern  Geometric series: 1 + r + r 2 + … + r n = (r n+1 – 1)/(r – 1)  Arithmetic series: … + n = n(n+ 1)/2

 Use the method of iteration to find an explicit formula for the following recursively defined sequence:  d k = 2d k−1 + 3, for all integers k ≥ 2  d 1 = 2  Use a proof by induction to show that your explicit formula is correct

 To solve sequence a k = Aa k-1 + Ba k-2  Find its characteristic equation t 2 – At – B = 0  If the equation has two distinct roots r and s  Substitute a 0 and a 1 into a n = Cr n + Ds n to find C and D  If the equation has a single root r  Substitute a 0 and a 1 into a n = Cr n + Dnr n to find C and D

 Find an explicit formula for the following:  r k = 2r k-1 − r k-2, for all integers k ≥ 2  r 0 = 1  r 1 = 4

 Defining finite and infinite sets  Definitions of:  Subset  Proper subset  Set equality  Set operations:  Union  Intersection  Difference  Complement  The empty set  Partitions  Cartesian product

 Proving a subset relation  Element method: Assume an element is in one set and show that it must be in the other set  Algebraic laws of set theory: Using the algebraic laws of set theory (given on the next slide), we can show that two sets are equal  Disproving a universal statement requires a counterexample with specific sets

NameLawDual Commutative A  B = B  AA  B = B  A Associative (A  B)  C = A  (B  C)(A  B)  C = A  (B  C) Distributive A  (B  C) = (A  B)  (A  C)A  (B  C) = (A  B)  (A  C) Identity A   = AA  U = A Complement A  A c = UA  A c =  Double Complement(A c ) c = A Idempotent A  A = AA  A = A Universal Bound A  U = UA   =  De Morgan’s (A  B) c = A c  B c (A  B) c = A c  B c Absorption A  (A  B) = AA  (A  B) = A Complements of U and  U c =  c = U Set Difference A – B = A  B c

 It is possible to give a description for a set which describes a set that does not actually exist  For a well-defined set, we should be able to say whether or not a given element is or is not a member  If we can find an element that must be in a specific set and must not be in a specific set, that set is not well defined

 Definitions  Domain  Co-domain  Range  Inverse image  Arrow diagrams  Poorly defined functions  Function equality

 One-to-one (injective) functions  Onto (surjective) functions  If a function F: X  Y is both one-to-one and onto (bijective), then there is an inverse function F -1 : Y  X such that:  F -1 (y) = x  F(x) = y, for all x  X and y  Y

 Pigeonhole principle:  If n pigeons fly into m pigeonholes, where n > m, then there is at least one pigeonhole with two or more pigeons in it  Cardinality is the number of things in a set  It is reflexive, symmetric, and transitive  Two sets have the same cardinality if a bijective function maps every element in one to an element in the other  Any set with the same cardinality as positive integers is called countably infinite

 Relations are generalizations of functions  In a relation (unlike functions), an element from one set can be related to any number (from zero up to infinity) of other elements  We can define any binary relation between sets A and B as a subset of A x B  If x is related to y by relation R, we write x R y  All relations have inverses (just reverse the order of the ordered pairs)

 For relation R on set A  R is reflexive iff for all x  A, (x, x)  R  R is symmetric iff for all x, y  A, if (x, y)  R then (y, x)  R  R is transitive iff for all x, y, z  A, if (x, y)  R and (y, z)  R then (x, z)  R  R is antisymmetric iff for all a and b in A, if a R b and b R a, then a = b  The transitive closure of R called R t satisfies the following properties:  R t is transitive R  RtR  Rt  If S is any other transitive relation that contains R, then R t  S

 Let A be partitioned by relation R  R is reflexive, symmetric, and transitive iff it induces a partition on A  We call a relation with these three properties an equivalence relation  Example: congruence mod 3  If R is reflexive, antisymmetric, and transitive, it is called a partial order  Example: less than or equal

 Prove that the subset relationship is a partial order  Consider the relation x R y, where R is defined over the set of all people  x R y ↔ x lives in the same house as y  Is R an equivalence relation? Prove it.

 A sample space is the set of all possible outcomes  An event is a subset of the sample space  Formula for equally likely probabilities:  Let S be a finite sample space in which all outcomes are equally likely and E is an event in S  Let N(X) be the number of elements in set X ▪ Many people use the notation |X| instead  The probability of E is P(E) = N(E)/N(S)

 If an operation has k steps such that  Step 1 can be performed in n 1 ways  Step 2 can be performed in n 2 ways …  Step k can be performed in n k ways  Then, the entire operation can be performed in n 1 n 2 … n k ways  This rule only applies when each step always takes the same number of ways  If each step does not take the same number of ways, you may need to draw a possibility tree

 If a finite set A equals the union of k distinct mutually disjoint subsets A 1, A 2, … A k, then: N(A) = N(A 1 ) + N(A 2 ) + … + N(A k )  If A, B, C are any finite sets, then: N(A  B) = N(A) + N(B) – N(A  B)  And: N(A  B  C) = N(A) + N(B) + N(C) – N(A  B) – N(A  C) – N(B  C) + N(A  B  C)

 This is a quick reminder of all the different ways you can count k things drawn from a total of n things:  Recall that P(n,k) = n!/(n – k)!  And = n!/((n – k)!k!) Order MattersOrder Doesn't Matter Repetition Allowed nknk Repetition Not Allowed P(n,k)P(n,k)

 The binomial theorem states:  You can easily compute these coefficients using Pascal's triangle for small values of n

 Let A and B be events in the sample space S  0 ≤ P(A) ≤ 1  P(  ) = 0 and P(S) = 1  If A  B = , then P(A  B) = P(A) + P(B)  It is clear then that P(A c ) = 1 – P(A)  More generally, P(A  B) = P(A) + P(B) – P(A  B)

 Expected value is one of the most important concepts in probability, especially if you want to gamble  The expected value is simply the sum of all events, weighted by their probabilities  If you have n outcomes with real number values a 1, a 2, a 3, … a n, each of which has probability p 1, p 2, p 3, … p n, then the expected value is:

 Given that some event A has happened, the probability that some event B will happen is called conditional probability  This probability is:

 Review third third of the course

 Review chapters 10 – 12 and notes on grammars and automata