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Week 15 - Wednesday
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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?
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To prove x  D  P(x) we need to find at
least one element of D that makes P(x) true
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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
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1.
2.
3.
Express the statement to be proved in the form x  D, if P(x)
then Q(x)
Suppose that x is some specific (but arbitrarily chosen)
element of D for which P(x) is true
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
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If n is an integer, then:
 n is even   k  Z  n = 2k
 n is odd   k  Z  n = 2k + 1
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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
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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
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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
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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
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Unique factorization theorem: For any integer
n > 1, there exist a positive integer k, distinct
prime numbers p1, p2, …, pk, and positive
integers e1, e2, …, ek such that
e1 e2 e3
ek
1
2
3
k
Quotient remainder theorem: For any integer n
and any positive integer d, there exist unique
integers q and r such that
n  p p p ...p
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 n = dq + r and 0 ≤ r < d
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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
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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:
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 2, 5, 10, 17, 26, …
 ai = i2 + 1, i ≥ 1
Summation notation is used to describe a summation of some
part of a series
n
a

k m
k
 am  am 1  am 2  ...  an
Product notation is used to describe a product of some part of a
series
n
a
k m
k
 am  am 1  am 2  ...  an
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Student Lecture:
 Explain how a proof by mathematical induction
works
 Then use a proof by induction to prove the
following:
 For all integers n ≥ 3,43 + 44 + 45 + ⋯ + 4𝑛 =
4(4𝑛 −16)
3
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To prove a statement of the following form:
 n  Z, where n a, property P(n) is true
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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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Write the following in closed form:
 1+

1
2
1
+ 2
2
+
1
23
1
+⋯ 𝑛
2
Use mathematical induction to prove:
 For all integers n ≥ 1, 2 + 4 + 6+· · ·+2n = n2 + n
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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 + r2+ … + rn = (rn+1 – 1)/(r – 1)
 Arithmetic series: 1 + 2 + 3 + … + n = n(n+ 1)/2
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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
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Use a proof by induction to show that your
explicit formula is correct
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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
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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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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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Defining finite and infinite sets
Definitions of:
 Subset
 Proper subset
 Set equality
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Set operations:
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
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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
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Disproving a universal statement requires a
counterexample with specific sets
Name
Law
Dual
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 = 
Commutative
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
Uc = 
c = U
Complements of U and 
Set Difference
A – B = A  Bc
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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
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Definitions
 Domain
 Co-domain
 Range
 Inverse image
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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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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
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Cardinality is the number of things in a set
 It is reflexive, symmetric, and transitive
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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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Consider the set of integer complex numbers,
defined as numbers a + bi, where a, b ∈ Z and
i is −1
Prove that the set of integer complex
numbers is countable
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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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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
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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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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
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If R is reflexive, antisymmetric, and
transitive, it is called a partial order
 Example: less than or equal
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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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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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If an operation has k steps such that
 Step 1 can be performed in n1 ways
 Step 2 can be performed in n2 ways
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…
 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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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)
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If A, B, C are any finite sets, then:
N(A  B) = N(A) + N(B) – N(A  B)
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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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This is a quick reminder of all the different ways
you can count k things drawn from a total of n
things:
Order Matters
Order Doesn't Matter
Repetition Allowed
nk
k  n 1


 k 
Repetition Not Allowed
P(n,k)
n
 
k 
Recall that P(n,k) = n!/(n – k)!
And  n  = n!/((n – k)!k!)
k 
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The binomial theorem states:
n

 n k k
n
(a  b)    a b
k 0  k 
n

You can easily compute these coefficients
using Pascal's triangle for small values of n
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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)
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
a p
k 1
k
k
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Given that some event A has happened, the
probability that some event B will happen is
called conditional probability
This probability is:
P ( A  B)
P(B | A) 
P( A)
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Review third third of the course
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Review chapters 10 – 12 and notes on
grammars and automata