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Week 15 - Wednesday.  What did we talk about last time?  Review first third of course.

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Presentation on theme: "Week 15 - Wednesday.  What did we talk about last time?  Review first third of course."— Presentation transcript:

1 Week 15 - Wednesday

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

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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?

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6  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

7  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

8  Use a proof by contradiction to prove the following:  For all integers n, if n 2 is odd then n is odd  For all prime numbers a, b, and c, a 2 + b 2 ≠ c 2

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10  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

11  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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13  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

14  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

15  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

16  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

17  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

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

20  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

21 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

22  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

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

24  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

25  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

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28  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)

29  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

30  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

31  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.

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33  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)

34  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

35  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)

36  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)

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

38  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)

39  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:

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

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43  Review third third of the course

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


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