desmath(1).ppt

Course Objectives:
• 1-Develop basic mathematical reasoning
skills for understanding and constructing
• mathematical arguments.
• 2. Learn how to analyze and solve counting
problems.
• 3. Learn how to specify, verify, and analyze
simple algorithms

Topics Covered:
1. Formal logic
2. Proofs
4. Sets and combinatorics
5. Algorithms and their complexity
6. Induction
7. Counting
8. Discrete probability
9. Relations
10. Graphs and trees
11. Boolean algebra
12. Finite-state machines
13. Turing machines

Textbooks:
• 1 Rosen, “Discrete Mathematics and Its
Applications”, 6th ed., McGraw-Hill,.
•

Chapter 1
Propositional Logic
Formal logic

1.1 Propositional Logic
A proposition :is a declarative sentence (that is, a sentence that declares a
fact) that is either true or false, but not both.
EXAMPLE 1: l All the following declarative sentences are propositions.
1. Washington., is the capital of the USA.
2 . Omdurman is the capital of Sudan .
3 . 1 + 1 = 2 .
4 . 2 + 2 = 3 .
Propositions 1 and 3 are true, whereas 2 and 4 are false
EXAMPLE 2: Consider the following sentences.
1 . What time is it?
2 . Read this carefully.
3 . x + 1 = 2 .
4 . x + y = Z

Sentences 1 and 2 are not propositions because they are not
declarative sentences. Sentences 3 and 4 are not propositions
because they are neither true nor false. Note that each of sentences 3
and 4 can be turned into a proposition if we assign values to the
variables.
We use letters to denote propositional variables (or statement
variables), that is, variables that represent propositions, just as letters
are used to denote numerical variables.

cont
The conventional letters used for propositional
variables are p, q , r, s, . . . . The truth value of a
proposition is true, denoted by T, if it is a true
proposition and false, denoted by F, if it is a false
proposition.
The area of logic that deals with propositions is called
the propositional calculus or propositional logic.
It was first developed systematically by the Greek
philosopher Aristotle more than 300 years ago

Compound: propositions, are formed from existing
propositions using logical operators.
DEFINITION 1 :Let p be a proposition. The negation of p, denoted
by p, is the statement
"It is not the case that p."
The proposition p is read "not p." The truth value of the
negation of p, p, is the opposite
of the truth value of p.

• EXAMPLE 3: Find the negation of the proposition
• "Today is Friday."
• Solution: The negation is
• "It is not the case that today is Friday."
• This negation can be more simply expressed by
• "Today is not Friday,"
• Or, "It is not Friday today."

DEFINITION 2:Let p and q be propositions. The
conjunction of p and q, denoted by p  q, is the
proposition "p and q ." The conjunction p  q is
true when both p and q are true and is false
otherwise.
EXAMPLE 4: Find the conjunction of the
propositions p and q where p is the proposition
"Today is Friday"
and q is the proposition "It is raining today."

• Solution: The conjunction of these propositions,
p  q, is the proposition "Today is Friday and
• it is raining today." This proposition is true on
rainy Fridays and is false on any day that is not a
Friday and on Fridays when it does not .rain

DEFINITION 3:Let p and q be propositions. The
disjunction of p and q, denoted by p V q, is the
proposition
"p or q ." The disjunction p V q is false when both p
and q are false and is true otherwise.

• The use of the connective or in a disjunction
corresponds to one of the two ways the word
• or is used in English, namely, in an inclusive
way. A disjunction is true when at least one of
the two propositions is true. For instance, the
inclusive or is being used in the statement
• "Students who have taken calculus or
computer science can take this class.

EXAMPLE 5: What is the disjunction of the
propositions p and q where p and q are the same
propositions as
in Example 4?
Solution: The disjunction of p and q, p v q, is the
proposition
"Today is Friday or it is raining today."
This proposition is true on any day that is either a
Friday or a rainy day (including rainy Fridays).
It is only false on days that are not Fridays when it
also does not rain.

DEFINITION 4: Let p and q be propositions. The
exclusive or of p and q, denoted by p q (p XOR q),
is the proposition that is true when exactly one of p
and q is true and is false otherwise.
p q p ⊕ q
T T F
T F T
F T T
F F F

Conditional Statements:
Let p and q be propositions. The conditional statement p → q is the proposition "if p,
then q ." The conditional statement p → q is false when p is true and q is false, and true
otherwise.
In the conditional statement p → q , p is called the hypothesis (premise) and q is called
the conclusion (or consequence).
The statement p→ q is called a conditional statement because p → q asserts that q is
true on the condition that p holds. A conditional statement is also called an implication.
The truth table for the conditional statement p → q is shown in Table 5. Note that the
statement p → q is true when both p and q are true and when p is false (no matter what
truth value q has).

The statement p→ q is called a conditional statement because p → q asserts that q is
true on the condition that p holds. A conditional statement is also called an implication.
The truth table for the conditional statement p → q is shown in Table 5. Note that the
statement p → q is true when both p and q are true and when p is false (no matter what
truth value q has).
ways to express this conditional statement:
"if p, then q"
"if p, q "
"p is sufficient for q"
"q if p"
"q when p"
"a necessary condition for p is q"
"q unless → p"
"p implies q"
"p only if q"
"a sufficient condition for q is p"
"q whenever p"
"q is necessary for p"
"q follows from p"

CONVERSE, CONTRAPOSITIVE, AND INVERSE We
can form some new conditional statements
starting with a conditional statement p → q . In
particular, there are three related conditional.
The proposition q → p is called the converse of p
→ q
The contra positive of p → q is the proposition ¬ q
→ ¬ p.
The proposition ¬ p → ¬ q is called the inverse of
p → q.

EXAMPLE 9: What are the contrapositive, the
converse, and the inverse of the conditional
statement
"The home team wins whenever it is raining."?
Solution: Because "q whenever p" is one of the
ways to express the conditional statement p → q ,
the original statement can be rewritten as
"If it is raining, then the home team wins."
Consequently, the contrapositive of this
conditional statement is "If the home team does
not win, then it is not raining."

The converse is
"If the home team wins, then it is raining."
The inverse is
"If it is not raining, then the home team does not win."
Only the contrapositive is equivalent to the original
statement.
BICONDITIONALS: another way to combine propositions
that expresses that two propositions have the same truth
value.

DEFINITION 6 :Let p and q be propositions. The biconditional
statement p q is the proposition "p if and only if q ." The
biconditional statement p q is true when p and q have the same
truth values, and is false otherwise. Biconditional statements are
also called bi-implications.

EXAMPLE 10: Let p be the statement "You can
take the flight" and let q be the statement "You
buy a ticket."
Then p q is the statement
"You can take the flight if and only if you buy a
ticket."
This statement is true if p and q are either both
true or both false, that is, if you buy a ticket and
can take the flight or if you do not buy a ticket
and you cannot take the flight. It is false when p
and q have opposite truth values.

EXAMPLE 11: Construct the truth table of the
compound proposition
(p v ¬ q) → (p  q).

EXAMPLE 12 :How can this English sentence be
translated into a logical expression? "You can
access the Internet from campus only if you are a
computer science major or you are not a
freshman.” a= access the internet,
C = computer science student and,
F =freshman

Chapter 1.2
Propositional Equivalences

1.2 Propositional Equivalences
DEFINITION 1:A compound proposition that is
always true, no matter what the truth values of the
propositions that occur in it, is called a tautology. A
compound proposition that is always false is called
a contradiction. A compound proposition that is
neither a tautology nor a contradiction is called a
contingency. Tautologies and contradictions are
often important in mathematical reasoning.

EXAMPLE 1: We can construct examples of tautologies and
contradictions using just one propositional variable.
Logical Equivalences: Compound propositions that have the same
truth values in all possible cases are called logically equivalent. We
can also define this notion as follows.
DEFINITION 2: The compound propositions p and q are called
logically equivalent if p q is a tautology.
The notation p == q denotes that p and q are logically equivalent.

EXAMPLE 4 :Show that
are logically equivalent. This is the distributive law of
disjunction over conjunction.

Propositional operations summary

Bit Operations
Boolean values can be represented as 1 (true) and 0 (false)
A bit string is a series of Boolean values. Length of the string
is the number of bits.
10110100 is eight Boolean values in one string
 We can then do operations on these Boolean strings.
Each column is its own Boolean operation
01011010
10110100
11101110

Bit Operations
EXAMPLE : Find the bitwise OR, bitwise AND, and bitwise XOR of the bit strings
0 1 1 0 1 1 0 1 1 0 and 1 1 000 1 1 1 0 1 . (Note that: bit strings will be split into
blocks of four bits to make them easier to read.)
Solution: The bitwise OR, bitwise AND, and bitwise XOR of these strings are
obtained by taking the OR, AND, and XOR of the corresponding bits, respectively.
0 1 1 0 1 1 0 1 1 0
1 1 0 0 0 1 1 1 0 1
1 1 1 0 1 1 1 1 1 1 bitwise OR
0 1 0 0 0 1 0 1 0 0 bitwise AND
1 0 1 0 1 0 1 0 1 1 bitwise XOR

Proof using Logical Equivalence
(p  r)  (q  r)
 ( p  r)  ( q  r) Definition of implication
  p  r   q  r Associative
  p   q  r  r Commutative
( p   q)  (r  r) Associative
  (p  q)  r De Morgan, Idempotent
 (p  q)  r Definition of implication

Example
Prove that (p  q)  (p  q) is a Tautology.
(Proof)
(p  q)  (p  q)
      (p  q)  (p  q) Implication
 ( p   q)  (p  q) De Morgan
( p  p)  ( q  q) Commutative, Associative
T  T Negation
 T Identity

Tautology, Contradiction, Equivalence
 Tautology: a statement that’s always true
p   p will always be true
 Contradiction: a statement that’s always false
p   p will always be false
A logical equivalence means that the two sides always have the same
truth values
Symbol is ≡ or  (we’ll use ≡)

Homework 1
Question 7 from Rosen, p. 17
p = “It is below freezing”
q = “It is snowing”
• It is below freezing and it is snowing
• It is below freezing but not snowing
• It is not below freezing and it is not snowing
• It is either snowing or below freezing (or both)
• If it is below freezing, it is also snowing
• It is either below freezing or it is snowing,
but it is not snowing if it is below freezing
• That it is below freezing is necessary and
sufficient for it to be snowing

Chapter 1.3
Predicates and Quantifiers

1.3 Predicates and Quantifiers
 How can we express?
“every computer in CS department is protected by
intrusion detection system”
“There exists at least one student who has a red hair”.
 “x is greater than 3”
x: subject
“is greater than 3”: predicate
P(x): propositional function P at x

Propositional Functions
Consider P(x) = x < 5
P(x) has no truth values (x is not given a value)
P(1) is true: The proposition 1<5 is true
P(10) is false: The proposition 10<5 is false
P(x) will create a proposition when given a value
Let P(x) = “x is a multiple of 5”
For what values of x is P(x) true?
Let P(x) = x + 3
For what values of x is P(x) true?

Functions with multiple variables:
• P(x, y) = x + y == 0
P(1,2) is false, P(1,-1) is true
• P(x, y, z) = x + y == z
P(3,4,5) is false, P(1,2,3) is true
• P(x1,x2,x3 … xn) = …

EXAMPLE 1 :

EXAMPLE 3:

Quantifiers
 Why quantifiers?
Many things (in this course and beyond) are specified using
quantifiers
A quantifier is “an operator that limits the variables of a
proposition”
Two types:
1- Universal
2- Existential

Universal quantifiers
DEFINITION 1:
The universal quantification of P (x) is the statement
"P(x) for all values of x in the domain."
The notation x P (x) denotes the universal quantification
of P(x). Here is called the universal quantifier.
We read x P(x) as "for all x P(x)" or "for every x P(x )."
An element for which P(x) is false is called a counterexample
of x P (x).

Represented by an upside-down A: 
• It means “for all”
• Let P(x) = x+1 > x
We can state the following:
• x P(x)
• English translation: “for all values of x, P(x) is true”
• English translation: “for all values of x, x+1>x is true”
But is that always true?
x P(x)
Let x = the character ‘a’
Is ‘a’+1 > ‘a’?

You need to specify your universe!
• What values x can represent
• Called the “domain” or “universe of discourse” by the
textbook.
Let the universe be the real numbers.
Let P(x) = x/2 < x
• Not true for the negative numbers!
• Thus, x P(x) is false
When the domain is all the real numbers
In order to prove that a universal quantification is true, it
must be shown for ALL cases
In order to prove that a universal quantification is false, it
must be shown to be false for only ONE case

Existential quantification
 Represented by an backwards E: 
• It means “there exists”
• Let P(x) = x+1 > x
 We can state the following:
• x P(x)
• English translation: “there exists (a value of) x
such that P(x) is true”
• English translation: “for at least one value of x,
x+1>x is true”

DEFINITION 2 :The existential quantification of P(x)
• Note that you still have to specify your universe
• Let P(x) = x+1 < x
There is no numerical value x for which
x+1<x
Thus, x P(x) is false

• Let P(x) = x+1 > x
There is a numerical value for which x+1>x
In fact, it’s true for all of the values of x!
Thus,  x P(x) is true
• In order to show an existential quantification is
true, you only have to find ONE value.
• In order to show an existential quantification is
false, you have to show it’s false for ALL values

EXAMPLE 14:

A note on quantifiers
• Recall that P(x) is a propositional function
Let P(x) be “x == 0”
• Recall that a proposition is a statement that is either true or false
 P(x) is not a proposition
• There are two ways to make a propositional function into a
proposition:
 Supply it with a value
For example, P(5) is false, P(0) is true
 Provide a quantification
For example, x P(x) is false and x P(x) is true
Let the universe of discourse be the real numbers

Binding variables
 Let P(x, y) be x > y
• Consider: x P(x, y)
This is not a proposition!
• What is y?
If it’s 5, then x P(x, y) is false
If it’s x-1, then x P(x, y) is true
• Note that y is not “bound” by a quantifier

Binding variables
 (x P(x))  Q(x)
The x in Q(x) is not bound; thus not a proposition
 (x P(x))  (x Q(x))
Both x values are bound; thus it is a proposition
 (x P(x)  Q(x))  (y R(y))
All variables are bound; thus it is a proposition
 (x P(x)  Q(y))  (y R(y))
The y in Q(y) is not bound; this not a proposition

Negating quantifications
 Consider the statement:
• All students in this class have red hair
 What is required to show the statement is false?
• There exists a student in this class that does NOT have
red hair
 To negate a universal quantification:
• You negate the propositional function
• AND you change to an existential quantification
• ¬x P(x) = x ¬P(x)

Negating quantifications
Consider the statement:
• There is a student in this class with red hair
 What is required to show the statement is false?
• All students in this class do not have red hair
 Thus, to negate an existential quantification:
• To negate the propositional function
• AND you change to a universal quantification
• ¬x P(x) = x ¬P(x)

Chapter 1.4
Nested Quantifiers

1.4 Nested Quantifiers
EXAMPLE 1 :

Multiple quantifiers
 You can have multiple quantifiers on a statement
 x y P(x, y)
“For all x, there exists a y such that P(x, y)”
Example: x y (x + y == 0)
 x y P(x, y)
There exists an x such that for all y P(x, y) is true”
x y (x*y == 0)

Order of quantifiers
 x y and x y are not equivalent!
 x y P(x, y)
P(x, y) = (x + y == 0) is false
 x y P(x, y)
P(x, y) = (x + y == 0) is true

Negating multiple quantifiers
 Recall negation rules for single quantifiers:
• ¬x P(x) = x ¬P(x)
• ¬x P(x) = x ¬P(x)
• Essentially, you change the quantifier(s), and negate
what it’s quantifying
 Examples:
• ¬(x y P(x, y)) = x ¬y P(x, y) = x y ¬P(x, y)
• ¬(x y z P(x, y, z)) = x ¬y z P(x, y, z)
= x y ¬ z P(x, y, z) = x y z ¬P(x, y, z)

Negating multiple quantifiers
 Consider ¬(x y P(x, y)) = x y ¬P(x, y)
• The left side is saying “for all x, there exists a y such
that P is true”
• To disprove it (negate it), you need to show that “there
exists an x such that for all y, P is false”
 Consider ¬(x y P(x, y)) = x y ¬P(x, y)
• The left side is saying “there exists an x such that for all
y, P is true”
• To disprove it (negate it), you need to show that “for all
x, there exists a y such that P is false”

Nested Quantifiers Summary
TABLE 1 :

Translating between English and quantifiers
 The product of two negative integers is positive
x y ((x<0)  (y<0) → (x y > 0))
• Why conditional instead of and?
 The average of two positive integers is positive
x y ((x>0)  (y>0) → ((x + y)/2 > 0))
 The difference of two negative integers is not necessarily
negative
x y ((x<0)  (y<0)  (|x-y|≥0))
• Why and instead of conditional?
 The absolute value of the sum of two integers does not
exceed the sum of the absolute values of these integers
x y (|x + y| ≤ |x| + |y|)

Translating between English and quantifiers
 x y (x + y = y)
There exists an additive identity for all real numbers
 x y (((x≥0)  (y<0)) → (x - y > 0))
A non-negative number minus a negative number is
greater than zero
 x y (((x≤0)  (y≤0))  (x-y > 0))
The difference between two non-positive numbers is not
necessarily non-positive (i.e. can be positive)
 x y (((x≠0)  (y≠0)) ↔ (x y ≠ 0))
The product of two non-zero numbers is non-zero if and
only if both factors are non-zero

Chapter 1.5
Rules of Inference

Valid Arguments
 Assume you are given the following two statements:
• "If you have a current password, then you can log
onto the network."
• "You have a current password."
• Therefore,
• "You can log onto the network."
p  q
p
 q

Definitions
 An Argument in propositional logic is a sequence of
propositions.
 All but the final proposition are called premises.
 The final proposition is called conclusion.
 An argument is valid if the truth of all premises implies that
the conclusion is true.
i.e. (p1  p2  …  pn)  q is a tautology.

Modus Ponens
 Consider (p  (p → q)) → q
p
p  q
 q

Modus Ponens example
“you are in this class”
“if you are in this class, you will get a grade”
 Let p = “you are in this class”
 Let q = “you will get a grade”
 By Modus Ponens, you can conclude that you will get a
grade

Modus Tollens
 Assume that we know: ¬q and p → q
• Recall that p → q  ¬q → ¬p
 Thus, we know ¬q and ¬q → ¬p
 We can conclude ¬p
 q
p  q
  p

Modus Tollens example
“you will not get a grade”
“if you are in this class, you will get a grade”
 Let p = “you are in this class”
 Let q = “you will get a grade”
 By Modus Tollens, you can conclude that you are not in this
class

Addition & Simplification
 Addition: If you know that p is true, then p  q will
ALWAYS be true
p
 p  q
 Simplification: If p  q is true, then p will ALWAYS be
true
p  q
 q

Example Proof
 We have the hypotheses:
• “It is not sunny this afternoon and it is colder than yesterday”
• “We will go swimming only if it is sunny”
• “If we do not go swimming, then we will take a canoe trip”
• “If we take a canoe trip, then we will be home by sunset”
 Does this imply that “we will be home by sunset”?
 (( p  q)  (r  p)  ( r  s)  (s  t))  t ???
• When
 p = “It is sunny this afternoon”
 q = “it is colder than yesterday”
 r = “We will go swimming”
 s = “we will take a canoe trip”
 t = “we will be home by sunset

Example of proof
1. ¬p  q 1st hypothesis
2. ¬p Simplification using step 1
3. r → p 2nd hypothesis
4. ¬r Modus tollens using steps 2 & 3
5. ¬r → s 3rd hypothesis
6. s Modus ponens using steps 4 & 5
7. s → t 4th hypothesis
8. t Modus ponens using steps 6 & 7

More Rules of Inference
 Conjunction: if p and q are true separately, then pq is
true
 Disjunctive syllogism: If pq is true, and p is false, then q
must be true
 Resolution: If pq is true, and ¬pr is true, then qr must
be true
 Hypothetical syllogism: If p→q is true, and q→r is true,
then p→r must be true

1.5 Rules of Inference Summary
.

1.5 Rules of Inference
Definition:
EXAMPLE 1 :Suppose that the conditional statement "If it snows today,
then we will go skiing" and its hypothesis, "It is snowing today," are true.
Then, by modus ponens, it follows that the conclusion of the conditional
statement, "We will go skiing," is true..

1.5 Rules of Inference
EXAMPLE 5

Chapter 1.6
Introduction to Proofs

Terminology
 Theorem: a statement that can be shown true. Sometimes
called facts.
• Proposition: less important theorem
 Proof: Demonstration that a theorem is true.
 Axiom: A statement that is assumed to be true.
 Lemma: a less important theorem that is useful to prove a
theorem.
 Corollary: a theorem that can be proven directly from a
theorem that has been proved.
 Conjecture: a statement that is being proposed to be a true
statement.

Terminology
Rules of inference, together with definitions of terms, are
used to draw conclusions from other assertions, tying
together the steps of a proof.
 In practice, the final step of a proof is usually just the
conclusion of the theorem. However, for clarity, we will
often recap the statement of the theorem as the final step of
a proof.
DEFINITION 1:The integer n is even if there exists an
integer k such that n = 2k, and n is odd if there exists an
integer k such that n = 2k + 1 . (Note that an integer is either
even or odd, and no integer is both even and odd.)

Proof methods
 We will discuss ten proof methods:
1. Direct proofs
2. Indirect proofs
3. Vacuous proofs
4. Trivial proofs
5. Proof by contradiction
6. Proof by cases
7. Proofs of equivalence
8. Existence proofs
9. Uniqueness proofs
10.Counterexamples

Direct proofs
 Consider an implication: p → q
• If p is false, then the implication is always true
• Thus, show that if p is true, then q is true
 To perform a direct proof, assume that p is true, and show that q must
therefore be true
 Show that the square of an even number is an even number
• Rephrased: if n is even, then n2 is even
 (Proof) Assume n is even
Thus, n = 2k, for some k (definition of even numbers)
n2 = (2k)2 = 4k2 = 2(2k2)
As n2 is 2 times an integer, n2 is thus even

Indirect proofs
• It’s contrapositive is ¬q →¬p
 Is logically equivalent to the original implication!
• If the antecedent (¬q) is false, then the contrapositive is
always true
• Thus, show that if ¬q is true, then ¬p is true
 To perform an indirect proof, do a direct proof on the
contrapositive

Indirect proof example
 If n2 is an odd integer then n is an odd integer
 Prove the contrapositive: If n is an even integer, then n2 is
an even integer
 Proof: n=2k for some integer k (definition of even
numbers)
 n2 = (2k)2 = 4k2 = 2(2k2)
 Since n2 is 2 times an integer, it is even
 When do you use a direct proof versus an indirect proof?

Example of which to use
 Prove that if n is an integer and n3+5 is odd, then n is even
 Via direct proof
• n3+5 = 2k+1 for some integer k (definition of odd
numbers)
• n3 = 2k-4
• Umm… ???
 So direct proof didn’t work out. So: indirect proof
• Contrapositive: If n is odd, then n3+5 is even
• Assume n is odd, and show that n3+5 is even
• n=2k+1 for some integer k (definition of odd numbers)
• n3+5 = (2k+1)3+5 = 8k3+12k2+6k+6 = 2(4k3+6k2+3k+3)
• As 2(4k3+6k2+3k+3) is 2 times an integer, it is even
3
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Proof by contradiction
 Given a statement p, assume it is false
• Assume ¬p
 Prove that ¬p cannot occur
• A contradiction exists
 Given a statement of the form p → q
• To assume it’s false, you only have to consider the case
where p is true and q is false

Proof by contradiction example
 Prove that if n is an integer and n3+5 is odd, then n is even
 Rephrased: If n3+5 is odd, then n is even
 Assume p is true and q is false
• Assume that n3+5 is odd, and n is odd
 n=2k+1 for some integer k (definition of odd numbers)
 n3+5 = (2k+1)3+5 = 8k3+12k2+6k+6 = 2(4k3+6k2+3k+3)
 As 2(4k3+6k2+3k+3) is 2 times an integer, it must be even
 Contradiction!

Proof by contradiction example
 Theorem (by Euclid): There are infinitely many prime
numbers.
 Proof. Assume there are a finite number of primes
 List them as follows: p1, p2 …, pn.
 Consider the number q = p1p2 … pn + 1
• This number is not divisible by any of the listed primes
 If we divided pi into q, there would result a
remainder of 1
• We must conclude that q is a prime number, not among
the primes listed above
 This contradicts our assumption that all primes are in
the list p1, p2 …, pn.

Vacuous proofs
 If it can be shown that p is false, then the implication is
always true
• By definition of an implication
 Note that you are showing that the antecedent is false

Vacuous proof example
• All criminology majors in CS 2012 are female
• Rephrased: If you are a criminology major and you are in
CS 2012, then you are female
 Could also use quantifiers!
 Since there are no criminology majors in this class, the
antecedent is false, and the implication is true

Trivial proofs
 If it can be shown that q is true, then the implication is
always true
• By definition of an implication
 Note that you are showing that the conclusion is true
Trivial proof example
• If you are tall and are in CS 2012 then you are a student
 Since all people in CS 2012 are students, the implication is
true regardless

Proof by cases
Show a statement is true by showing all possible cases are
true
 Thus, you are showing a statement of the form:
(p1  p2  …  pn)  q
 is true by showing that:
[(p1p2…pn)q]  [(p1q)(p2q)…(pn q)]

Proof by cases example
 Prove that
Note that b ≠ 0
 Cases:
• Case 1: a ≥ 0 and b > 0
Then |a| = a, |b| = b, and
• Case 2: a ≥ 0 and b < 0
Then |a| = a, |b| = -b, and
• Case 3: a < 0 and b > 0
Then |a| = -a, |b| = b, and
• Case 4: a < 0 and b < 0
Then |a| = -a, |b| = -b, and
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Proofs of equivalences
This is showing the definition of a bi-conditional
 Given a statement of the form “p if and only if q”
• Show it is true by showing (p → q)(q → p) is true

Proofs of equivalence example
 Show that m2=n2 if and only if m=n or m=-n
 Rephrased: (m2=n2) ↔ [(m=n)(m=-n)]
[(m=n)(m=-n)] → (m2=n2)
• Proof by cases!
• Case 1: (m=n) → (m2=n2)
(m)2 = m2, and (n)2 = n2, so this case is proven
• Case 2: (m=-n) → (m2=n2)
(m)2 = m2, and (-n)2 = n2, so this case is proven
 (m2=n2) → [(m=n)(m=-n)]
• Subtract n2 from both sides to get m2-n2=0
• Factor to get (m + n)(m - n) = 0
• Since that equals zero, one of the factors must be zero
• Thus, either m + n=0 (which means m = - n)
• Or m - n=0 (which means m = n)

Existence proofs
 Given a statement: x P(x)
 We only have to show that a P(c) exists for some value of c
 Two types:
• Constructive: Find a specific value of c for which P(c) is
true.
• Non constructive: Show that such a c exists, but don’t
actually find it
 Assume it does not exist, and show a contradiction

Constructive existence proof example
 Show that a square exists that is the sum of two other
squares
• Proof: 32 + 42 = 52
 Show that a cube exists that is the sum of three other cubes
• Proof: 33 + 43 + 53 = 63

Non-constructive existence proof example
 Prove that either 2*10500+15 or 2*10500+16 is not a perfect
square
• A perfect square is a square of an integer
• Rephrased: Show that a non-perfect square exists in the
set {2*10500+15, 2*10500+16}
 Proof: The only two perfect squares that differ by 1 are 0
and 1
• Thus, any other numbers that differ by 1 cannot both be
perfect squares
• Thus, a non-perfect square must exist in any set that
contains two numbers that differ by 1
• Note that we didn’t specify which one it was!

Uniqueness proofs
 A theorem may state that only one such value exists
 To prove this, you need to show:
• Existence: that such a value does indeed exist
 Either via a constructive or non-constructive existence
proof
• Uniqueness: that there is only one such value

Uniqueness proof example
 If the real number equation 5x+3=a has a solution then it is
unique
 Existence
• We can manipulate 5x+3=a to yield x=(a-3)/5
• Is this constructive or non-constructive?
 Uniqueness
• If there are two such numbers, then they would fulfill the
following: a = 5x+3 = 5y+3
• We can manipulate this to yield that x = y
 Thus, the one solution is unique!

Counterexamples
 Given a universally quantified statement, find a single
example which it is not true
 Note that this is DISPROVING a UNIVERSAL statement
by a counterexample
 x ¬R(x), where R(x) means “x has red hair”
• Find one person (in the domain) who has red hair
 Every positive integer is the square of another integer
• The square root of 5 is 2.236, which is not an integer

What’s wrong with this proof?
 Theorem: All numbers are equal to 0.
(Proof)
Suppose a = b.
Then a2 = a b.
Then a2 - b2 = a b - b2.
Then (a - b)(a + b) = b(a - b).
Then a + b = b.
Therefore, a = 0.

desmath(1).ppt

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