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Published byAlfred Lang Modified over 8 years ago
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Week 8 - Wednesday
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What did we talk about last time? Relations Properties of relations Reflexive Symmetric Transitive
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There are five pirates who wish to divide 100 gold coins These pirates are ranked: 1. Captain 2. Lieutenant 3. Master 4. Midshipman 5. Seaman In order of rank, each pirate gets the opportunity to propose a plan for dividing up the gold If at least half of the pirates (including the proposer) agree on the proposition, it is carried out Otherwise, the pirate is killed and the next highest ranking pirate makes a proposal Pirates are completely rational, who value, in this order: 1. Staying alive 2. Maximizing gold coins received 3. Seeing other pirates die If you were the captain, what would you propose? Hint: Work backwards!
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If you have a set A and a binary relation R on A, the transitive closure of R called R t satisfies the following properties: R t is transitive R RtR Rt If S is any other transitive relation that contains R, then R t S Basically, the transitive closure just means adding in the least amount of stuff to R to make it transitive If you can get there in R, you can get there directly in R t
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Let A = {0, 1, 2, 3} Let R = {(0,1), (1,2), (2,3)} Is R transitive? Then, find the transitive closure of R
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Let R be a relation on real numbers R such that x R y x = y Is R reflexive? Is R symmetric? Is R transitive? Let S be a relation on real numbers R such that x S y x < y Is S reflexive? Is S symmetric? Is S transitive? Let T be a relation on positive integers N such that m T n 3 | (m – n) Is T reflexive? Is T symmetric? Is T transitive?
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A partition of a set A (as we discussed earlier) is a collection of nonempty, mutually disjoint sets, whose union is A A relation can be induced by a partition For example, let A = {0, 1, 2, 3, 4} Let A be partitioned into {0, 3, 4}, {1}, {2} The binary relation induced by the partition is: x R y x and y are in the same subset of the partition List the ordered pairs in R
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Given set A with a partition Let R be the relation induced by the partition Then, R is reflexive, symmetric, and transitive As it turns out, any relation R is that is reflexive, symmetric, and transitive induces a partition We call a relation with these three properties an equivalence relation
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We say that m is congruent to n modulo d if and only if d | (m – n) We write this: m n (mod d) Congruence mod d defines an equivalence relation Reflexive, because m m (mod d) Symmetric because m n (mod d) means that n m (mod d) Transitive because m n (mod d) and n k (mod d) mean that m k (mod d) Which of the following are true? 12 7 (mod 5) 6 -8 (mod 4) 3 3 (mod 7)
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Let A be a set and R be an equivalence relation on A For each element a in A, the equivalence class of a, written [a], is the set of all elements x in A such that a R x Example Let A be { 0, 1, 2, 3, 4, 5, 6, 7, 8} Let R be congruence mod 3 What's the equivalence class of 1? For A with R as an equivalence relation on A If b [a], then [a] = [b] If b [a], then [a] [b] =
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Modular arithmetic has many applications For those of you in Security, you know how many of them apply to cryptography To help us, the following statements for integers a, b, and n, with n > 1, are all equivalent 1. n | (a – b) 2. a b (mod n) 3. a = b + kn for some integer k 4. a and b have the same remainder when divided by n 5. a mod n = b mod n
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Let a, b, c, d and n be integers with n > 1 Let a c (mod n) and b d (mod n), then: 1. (a + b) (c + d) (mod n) 2. (a – b) (c – d) (mod n) 3. ab cd (mod n) 4. a m c m (mod n), for all positive integers m If a and n are relatively prime (share no common factors), then there is a multiplicative inverse a -1 such that a -1 a 1 (mod n) I'd love to have us learn how to find this, but there isn't time
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Let R be a relation on a set A R is antisymmetric iff for all a and b in A, if a R b and b R a, then a = b That is, if two different elements are related to each other, then the relation is not antisymmetric Let R be the "divides" relation on the set of all positive integers Is R antisymmetric? Let S be the "divides" relation on the set of all integers Is S antisymmetric?
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A relation that is reflexive, antisymmetric, and transitive is called a partial order The subset relation is a partial order Show it's reflexive Show it's antisymmetric Show it's transitive The less than or equal to relation is a partial order Show it's reflexive Show it's antisymmetric Show it's transitive
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In a proof by contradiction, you begin by assuming the negation of the conclusion Then, you show that doing so leads to a logical impossibility Thus, the assumption must be false and the conclusion true
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A proof by contradiction is different from a direct proof because you are trying to get to a point where things don't make sense You should always mark such proofs clearly Start your proof with the words Proof by contradiction Write Negation of conclusion as the justification for the negated conclusion Clearly mark the line when you have both p and ~p as a contradiction Finally, state the conclusion with its justification as the contradiction found before
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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 sequence Product notation is used to describe a product of some part of a sequence
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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
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Prove that, for all integers n 1,
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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
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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: 1 + 2 + 3 + … + n = n(n+ 1)/2
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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 There will be one of these on the exam
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Defining finite and infinite sets Definitions of: Subset Proper subset Set equality Set operations: Union Intersection Difference Complement The empty set Partitions Cartesian product
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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
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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
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Use the element method to prove the following: For all sets A, B, and C, if A ⊆ B then A ∩ C ⊆ B ∩ C
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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 Watch out for definitions that are logically inconsistent!
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Definitions Domain Co-domain Range Inverse image Arrow diagrams Poorly defined functions Function equality
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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
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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
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Exam 2!
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Finish Homework 6 Due at Exam 2 Study for Exam 2 Friday in class
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