Download presentation
Presentation is loading. Please wait.
Published byAnthony Melton Modified over 8 years ago
1
Week 15 - Wednesday
2
What did we talk about last time? Review first third of course
4
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?
6
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
7
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
8
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
9
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
10
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
12
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
14
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
16
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
17
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
18
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
19
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
20
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
22
Defining finite and infinite sets Definitions of: Subset Proper subset Set equality Set operations: Union Intersection Difference Complement The empty set Partitions Cartesian product
23
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
24
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
25
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
26
Definitions Domain Co-domain Range Inverse image Arrow diagrams Poorly defined functions Function equality
27
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
28
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
31
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)
32
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 RtR Rt If S is any other transitive relation that contains R, then R t S
33
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
34
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.
36
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)
37
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
38
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)
39
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)
40
The binomial theorem states: You can easily compute these coefficients using Pascal's triangle for small values of n
41
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)
42
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:
43
Given that some event A has happened, the probability that some event B will happen is called conditional probability This probability is:
46
Review third third of the course
47
Review chapters 10 – 12 and notes on grammars and automata
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.