week 15 - wednesday. what did we talk about last time? review first third of course
TRANSCRIPT
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CS322Week 15 - Wednesday
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Last time
What did we talk about last time? Review first third of course
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Questions?
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Logical warmup
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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Indirect Proofs
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Proof by contradiction
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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Contradiction formatting
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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Practice
Use a proof by contradiction to prove the following: For all integers n, if n2 is odd then n is
odd For all prime numbers a, b, and c, a2 +
b2 ≠ c2
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Sequences and Induction
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Sequences
Mathematical sequences can be represented in expanded form or with explicit formulas
Examples: 2, 5, 10, 17, 26, … ai = i2 + 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
nmmm
n
mkk aaaaa
...21
nmmm
n
mkk aaaaa
...21
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Proof by mathematical induction
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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Practice
Write the following in closed form:
Use mathematical induction to prove: For all integers n ≥ 1, 2 + 4 + 6+· · ·+2n
= n2 + n
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Recursion
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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Solving recursion by iteration
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 + r2+ … + rn =
(rn+1 – 1)/(r – 1) Arithmetic series: 1 + 2 + 3 + … + n =
n(n+ 1)/2
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Practice
Use the method of iteration to find an explicit formula for the following recursively defined sequence: dk = 2dk−1 + 3, for all integers k ≥ 2
d1 = 2 Use a proof by induction to show that
your explicit formula is correct
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Solving second order linear homogeneous recurrence relations with constant coefficients
To solve sequence ak = Aak-1 + Bak-2
Find its characteristic equation t2 – At – B = 0
If the equation has two distinct roots r and s Substitute a0 and a1 into an = Crn + Dsn
to find C and D If the equation has a single root r
Substitute a0 and a1 into an = Crn + Dnrn to find C and D
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Practice
Find an explicit formula for the following: rk = 2rk-1 − rk-2, for all integers k ≥ 2
r0 = 1
r1 = 4
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Set Theory
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Set theory basics
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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Set theory proofs
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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Laws of set theory
Name Law Dual
Commutative A B = B A A 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 = A A U = A
Complement A Ac = U A Ac =
Double Complement (Ac)c = A
Idempotent A A = A A A = A
Universal Bound A U = U A =
De Morgan’s (A B)c = Ac Bc (A B)c = Ac Bc
Absorption A (A B) = A A (A B) = A
Complements of U and
Uc = c = U
Set Difference A – B = A Bc
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Russell's paradox
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
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Functions
Definitions Domain Co-domain Range Inverse image
Arrow diagrams Poorly defined functions Function equality
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Inverses
One-to-one (injective) functionsOnto (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
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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Practice
Consider the set of integer complex numbers, defined as numbers a + bi, where a, b Z and i is
Prove that the set of integer complex numbers is countable
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Relations
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Relations
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)
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Properties of relations
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 Rt satisfies the
following properties: Rt is transitive R Rt
If S is any other transitive relation that contains R, then Rt S
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Equivalence relations and partial orders
Let A be partitioned by relation RR 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
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Practice
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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Counting and Probability
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Probability definitions
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)
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Multiplication rule
If an operation has k steps such that Step 1 can be performed in n1 ways
Step 2 can be performed in n2 ways
… Step k can be performed in nk ways
Then, the entire operation can be performed in n1n2 … nk 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
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Addition and inclusion/exclusion rules
If a finite set A equals the union of k distinct mutually disjoint subsets A1, A2, … Ak, then:N(A) = N(A1) + N(A2) + … + N(Ak)
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)
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Counting guide
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 nk
Repetition Not Allowed P(n,k)
k
nk 1
k
n
k
n
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Binomial theorem
The binomial theorem states:
You can easily compute these coefficients using Pascal's triangle for small values of n
kknn
k
n bak
nba
0
)(
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Probability axioms
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(Ac) = 1 – P(A) More generally, P(A B) = P(A) + P(B) –
P(A B)
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Expected value
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 a1, a2, a3, … an, each of which has probability p1, p2, p3, … pn, then the expected value is:
n
kkkpa
1
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Conditional probability
Given that some event A has happened, the probability that some event B will happen is called conditional probability
This probability is:
)()(
)|(AP
BAPABP
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Quiz
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Upcoming
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Next time…
Review third third of the course
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Reminders
Review chapters 10 – 12 and notes on grammars and automata