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Chapter One Notes
The Foundations: Logic and Proofs
Based on: Discrete Math & Its Applications - Kenneth Rosen
CSC125 - Spring 2010
As MS document
1.1 - Propositional Logic
Concepts::
Proposition
Propositional variables
Propositional logic (calculus)
Logical connectives (operators)      
Compound propositions
Logical operations
Negation
p
not p
Conjunction
pq
p and q
Disjunction
pq
p or q (or both)
Exclusive Or
pq
p xor q
Conditional (implication) p  q
if p, then q (p implies q)
Converse of p  q
is q  p (p  q)
The converse does not necessarily follow.
Example:
If it rains then the streets are wet – true
If the streets are wet then it rained – false
Contrapositive of p  q
is q  p
The contrapositive does necessarily follow.
If it rains then the streets are wet – true
If the streets are not wet then it did not rain – true
Inverse of p  q
is p  q
The inverse does not necessarily follow.
Example:
If it rains then the streets are wet – true
If it does not rain then the streets are not wet – false
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Biconditional p  q
if p, then q and if q, then p
p implies q and q implies p
p if and only if q
p iff q
Truth tables
p
p
T
F
F
T
p
q
pq
pq
pq pq pq
T
T
T
T
F
T
T
T
F
F
T
T
F
F
F
T
F
T
T
T
F
F
F
F
F
F
T
T
Bitwise And, Or, Xor
And 0
0
0
0
1
0
1
0
0
1
1
1
Or
0
0
0
0
1
1
1
0
1
1
1
1
Xor
0
0
0
0
1
1
1
0
1
1
1
0
2
How many Boolean operations?
x y f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15
0 0 0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0 1 0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
1 0 0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
1 1 0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
Natural language, logic and ambiguity
1.2 - Propositional Equivalences
Introduction::
From truth tables we can see that there are compound propositions that are equivalent
to each other. In other words, whenever the first proposition is true, so is the second.
And whenever the first is false, the second is also false. For example:
p
q
p
p q p  q
T
T
F
T
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
p
q
T
T
F
F
F
T
F
T
F
F
T
T
F
T
F
T
T
F
T
F
T
F
F
T
T
T
F
T
p q p  q p  q (p  q)
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Compare the above to Tables 3 and 4 on pp. 22 & 23 of Rosen. These are examples
of important equivalences which we will study. Before we do, we need to learn some
terminology.
Tautology – a proposition that is always true
Example: p  p
Contradiction (inconsistency) – a proposition that is always false
Example: p  p
Contingency – a proposition that is neither a tautology nor a contradiction, i.e., it is
sometimes true and sometimes false.
Example: p  q
For verification of these examples, look at Table 1, p. 22 and Table 7, p. 10.
Logical Equivalences::
p & q are logically equivalent if p ↔ q is a tautology
In that case we can write: p ≡ q
Truth tables can be used to establish whether or not p & q are logically equiv.
DeMorgan's Laws – Table 2 {on quiz }
Tables of important logical equivs. – Table 6 {on quiz}
Table 7 – involving conditionals {on quiz}
Table 8 – involving biconditionals {on quiz}
Constructing New Logical Equivalences::
Given a small set of logical equivalences, we can derive new ones in a step by step
process. Examples 6→8, pp. 26-27 illustrate how this is done. This process of
derivation of new propositions from old is very useful and forms the basis of a lot of
work in automated theorem proving.
1.3 - Predicates and Quantifiers
Predicate logic
predicate (propositional function)
variable
truth value of propositional function
n-place predicate or n-ary predicate
Quantifiers
quantification - expresses extent to which a predicate is true over a range of
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elements
predicate calculus
domain OR domain of discourse OR universe of discourse
universal quantifier
xP(x)
counterexample – one single x for which P(x) is false
existential quantifier
xP(x)
establishing example – one single x for which P(x) is true
See Table 1, p. 34
uniqueness quantifier
!xP(x) {or 1xP(x) }
quantifiers with restricted domains
precedence of quantifiers
xP(x)  Q(x) ≡ (xP(x))  Q(x) !≡ x(P(x))  Q(x))
binding variables
variable within scope of quantifier is bound
variable not within scope of quantifier is free
Logical equivalences
x(P(x) Q(x)) ≡xP(x) xQ(x), if same domain is used
throughout
Negating quantified expressions (DeMorgan's Laws)
x P(x) ≡ xP(x)
xP(x) ≡ xP(x)
Quantifiers related to  and 
xP(x) operates like a super 
Let domain be {1,2,3}
xP(x) ≡ P(1)  P(2)  P(3)
xP(x) operates like a super 
Let domain be {1,2,3}
xP(x) ≡ P(1)  P(2)  P(3)
That is why the negation laws above are called DeMorgan's Laws
x P(x) ≡ (P(1)  P(2)  P(3))
≡ P(1)  P(2)  P(3) ≡ xP(x)
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And
xP(x) ≡ (P(1)  P(2)  P(3))
≡ P(1) P(2)  P(3) ≡ xP(x)
See Table 2, p. 41
Translating to & from English
Let T(x) represent "x is tall"
Everyone is tall:: x T(x)
It is not true that everyone is tall:: x T(x), is the same as
Someone is not tall::
Someone is tall:: x T(x)
xT(x)
It is not true that someone is tall:: x T(x), is the same as
No one is tall:: xT(x), which is the same as
Everyone is not tall {although this is ambiguous; see below}
Ambiguity and subtlety of natural languages
It is important to realize that one of the difficulties in translating back and forth
from English (or any other natural language) to symbolic logic comes from the fact
that some expressions are used in more than one way, logically speaking. In addition
it is difficult to disambiguate because closely related expressions are used to express
subtle nuances of meaning.
Everyone is not tall is a good example. It can be taken to mean no one is tall,
as we took it to mean above.
xT(x)
Or it can be taken to mean not everyone is tall
x T(x)
or someone is not tall (which is logically equivalent)
x T(x)
That meaning comes out clearly in the following interchange.
Person A: Everyone is tall.
Person B: No, everyone is not tall.
Attempts to disambiguate by changing our wording lead to other problems. We
could try to express
xT(x)
as everyone is short, but that changes the meaning. If we have a basketball team
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where the heights range from 5'9" to 6'6", we can say everyone is not tall; but we
could not say everyone is short. We could make up new words and say everyone is
untall, but that is not likely to be welcomed by the general population of English
speakers.
It soon becomes pretty clear that there is no cleancut solution to the translation
problem. In fact, that is one of the reasons for using symbolic logic – to eliminate the
ambiguity inherent and widespread in natural language. Read the examples below to
see how prevalent ambiguity and subtlety are in our use of "the king's English."
Everyone goes
No, everyone does not
Everyone stays away
Everyone does not go
Everyone is going
Everyone is not going
Everyone is so not going
Someone goes
No, someone does not go
Some do not go
Some stay away
Everyone thinks
Everyone is unthinking
Everyone does not think
Everyone is thoughtful
Everyone is not thoughtful
Everyone is thoughtless
Someone thinks
Someone does not think
Some is thoughtless
Someone is thoughtless
Everyone is caring
Everyone is uncaring
Everyone is not caring
Not everyone is caring
Everyone is careless
Everyone is careful
Everyone is not careful
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Not everyone is careful
Not everyone is careless
Quick Summary
Realizing that there will still be ambiguous cases, here are some rules of thumb
x D(x) – everyone does
x D(x) – everyone does not; i.e., for everyone does not is true
x D(x) – not everyone does; i.e., for some does not is true
x D(x) – someone does
x D(x) – someone does not; there is someone who does not
x D(x) – no one does
Equivalences
x D(x) ≡ x D(x)
No one does is the same as everyone does not
x D(x) ≡ xD(x)
Not everyone does is the same as someone does not
1.4 - Nested Quantifiers
Think of quantification as loops
Example 2
Order of Quantifiers
Does not matter if all of same type
Example 3
Very important when of mixed type
Examples 4 & 5
Let domain be {1,2,3}
x y P(x,y) ≡ yx P(x,y)
≡ P(1,1)  P(1,2)  P(1,3)  P(2,1)  P(2,2)  P(2,3)  P(3,1) 
P(3,2)  P(3,3)
x y P(x,y) ≡ yx P(x,y)
≡ P(1,1)  P(1,2)  P(1,3)  P(2,1)  P(2,2)  P(2,3)  P(3,1) 
P(3,2)  P(3,3)
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x y P(x,y) ≡
(P(1,1)  P(1,2)  P(1,3)) 
(P(2,1)  P(2,2)  P(2,3)) 
(P(3,1)  P(3,2)  P(3,3))
In other words,
for x = 1, y P(x,y)  for x = 2, y P(x,y)  for x = 3, y P(x,y)
x y P(x,y) ≡
(P(1,1)  P(1,2)  P(1,3)) 
(P(2,1)  P(2,2)  P(2,3)) 
(P(3,1)  P(3,2)  P(3,3))
In other words,
either for x = 1, y P(x,y)  for x = 2, y P(x,y)  for x = 3, y P(x,y)
Similarly,
y x P(x,y) ≡
(P(1,1)  P(2,1)  P(3,1)) 
(P(1,2)  P(2,2)  P(3,2)) 
(P(1,3)  P(2,3)  P(3,3))
In other words,
either for y = 1, x P(x,y)  for y = 2, x P(x,y)  for y = 3, x P(x,y)
Therefore, x y P(x,y) ≠ y x P(x,y) ≠ x y P(x,y)
See Table 1, p. 53
Translating mathematical statements
Examples 6 & 7
Translating nested quantifiers into English
Examples 9 & 10
Translating English into logical expressions
Examples 11 & 13
Negating nested quantifiers
Example 14
x y (xy = 1) ≡
x  y (xy = 1) ≡
x y  (xy = 1) ≡
x y (xy ≠ 1)
9
Example 15
 w a f (P(w,f)  Q(f,a))
w a f (P(w,f)  Q(f,a))
w a f (P(w,f)  Q(f,a))
w a f (P(w,f)  Q(f,a))
w a f (P(w,f)  Q(f,a))
Doubly quantified expression examples:
Let W(x,y) be the statement “student x watched Eagles game y,” where the domain
for x consists of all students in this class and the domain for y consists of all
Philadelphia Eagles games last season. Express each proposition in English.
1. xyW(x,y)
A student from this class watched an Eagles game last season.
2. xyW(x,y)
No student from this class watched an Eagles game last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #18
{For all students and all Eagles games, it not true that any student
watched any game}
3. x yW(x,y)
A student from this class watched every Eagles game last season.
4. x yW(x,y)
No student from this class watched every Eagles game last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #16
5. yx W(x,y)
Every Eagles game last season was watched by a student in this class.
6. yx W(x,y)
10
Not every Eagles game last season was watched by a student in this
class.
Here are equivalent forms, by DeMorgan's Law
≡ yxW(x,y) ≡ yxW(x,y)
Note: Equivalent to #17
7. xyW(x,y)
Every student from this class watched an Eagles game last season.
8. xyW(x,y)
Not every student from this class watched an Eagles game last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #14
{A student from this class did not watch any Eagles games last season;
OR: It is not true that every student watched an Eagles game last
season}
9. yxW(x,y)
There is an Eagles game last season that every student watched.
10. yxW(x,y)
None of the Eagles games last season were watched by every student.
Here are equivalent forms, by DeMorgan's Law
≡ yxW(x,y) ≡ yxW(x,y)
Note: Equivalent to #15
11. x yW(x,y)
Every student from this class watched every Eagles game last season.
12. x yW(x,y)
Not every student from this class watched every Eagles game last
season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #13
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13. xyW(x,y)
A student from this class who missed an Eagles game last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #12
{There is a student from this class who did not watch every Eagles game
last season;
OR: Not every student from this class watched every Eagles game last
season.}
14. x yW(x,y)
A student from this class did not watch any Eagles games last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #8
{Not every student from this class watched an Eagles game last season;
OR: It is not true that every student watched an Eagles game last
season}
15. y x W(x,y)
No Eagles game was watched by every student in this class.
Here are equivalent forms, by DeMorgan's Law
≡ yxW(x,y) ≡ yxW(x,y)
Note: Equivalent to #10
16. xyW(x,y)
Every student from this class missed an Eagles game last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #4
{No student watched every Eagles game last season;
OR: For every student, there is an Eagles game they did not watch}
17. yxW(x,y)
There is an Eagles game that no one watched.
12
Here are equivalent forms, by DeMorgan's Law
≡ yxW(x,y) ≡ yxW(x,y)
Note: Equivalent to #6
18. xyW(x,y)
No student watched any Eagles game last season.
Here are equivalent forms, by DeMorgan's Law
≡ xyW(x,y) ≡ xyW(x,y)
Note: Equivalent to #2
Every student in this class missed watching every Eagles game last
season.
1.5 Rules of Inference
Valid arguments in propositional logic
p→q
p
.
q
Definitions:
argument - sequence of propositions
premises - all the propositions in the sequence except the final one
conclusion - the final proposition in the sequence
argument form - sequence involving propositional variables
valid - an argument form is valid if conclusion follows from premises, no
matter what values the variables have.
conclusion follows from premises - conclusion is true if all premises are true
Rules of inference for propositional logic
Study Table 1
Examples 1, 3, 4, 5
Using rules of inference to build arguments
Example 6
Hypotheses:
H1: p  q
H2: r  p
H3: r  s
H4: s  t
Argument:
13
1:
2:
3:
4:
5:
6:
7:
8:
p  q
p
rp
r
r  s
s
st
t
Example 7
Hypotheses:
H1: p  q
H2: p  r
H3: r  s
Argument:
1: p  q
2: q  p
3: p  r
4: q  r
5: r  s
6: q  s
H1
Simp (1)
H2
MT (2,3)
H3
MP (4,5)
H4
MP(6,7)
H1
Contrapos (1)
H2
HypSyl (2,3)
H3
HypSyl (4,5)
Resolution
((p  q)  (~p  r)) → (q  r)
Clause - disjunction of variables or their negations
Resolvent - conclusion of application of resolution rule
Fallacies
Affirming the conclusion
Denying the hypothesis
Rules of inference for quantified statements
Study Table 2
Examples 12 & 13
Example 12
Premises:
P1: x (D(x)  C(x))
P2: D(Marla)
Argument:
14
1:
2:
3:
4:
x (D(x)  C(x))
D(Marla)  C(Marla)
D(Marla)
C(Marla)
P1
UI (1)
P2
MP (2,3)
Example 13
Hypotheses:
P1: x (C(x)  B(x))
P2: x (C(x)  P(x))
Argument:
1: x (C(x)  B(x))
2: C(a)  B(a)
3: C(a)
4: x (C(x)  P(x))
5: C(a)  P(a)
6: P(a)
7: B(a)
8: P(a)  B(a)
9: x (P(x)  B(x))
P1
EI (1)
Simp (2)
P2
UI (4)
MP (3,5)
Simp (2)
Conj (6,7)
EG (8)
Combining rules of inference for propositions and quantified statements
Universal modus ponens
Universal modus tollens
Summary of Valid Rules of Inference (Valid Argument Forms)
Modus Ponens
p
p→q
q
Modus Tollens
q
p→q
 p
Hypothetical Syllogism
p→q
q→r
p→r
15
Disjunctive Syllogism
pq
p .
q
Addition
p
.
pq
Simplification
pq
p
Conjunction
p
q
pq
Resolution
pq
p  r
qr
Universal Instantiation
x P(x)
 P(c) for any element c
Universal Generalization
P(c) for an arbitrary c
 x P(x)
Existential Instantiation
x P(x)
 P(c) for some element c
Existential Generalization
P(c) for some element c
 x P(x)
Universal Modus Ponens
x (P(x) → Q(x))
16
P(a), where a is any element in the domain
 Q(a)
Universal Modus Tollens
x (P(x) → Q(x))
Q(a), where a is any element in the domain
 P(a)
Summary of Fallacies (Invalid Argument Forms)
Affirming the conclusion
q
p→q
p
Denying the hypothesis
p
p→q
 q
Unwarranted addition
p
.
pq
Unwarranted simplification
pq
p
Unwarranted Universal Generalization
P(c) for a particular c
 x P(x)
Existential Instantiation
x P(x)
 P(c) for a particular element c
Universal Affirming the conclusion
x (P(x) → Q(x))
Q(a), where a is any element in the domain
 P(a)
Universal Denying the Hypothesis
17
x (P(x) → Q(x))
P(a), where a is any element in the domain
 Q(a)
1.6 Introduction to Proofs
Terminology
Theorem – statement that can be proven to be true.
Propositions, facts, results – less important true statements.
Proof – valid argument that establishes truth of a theorem.
Axiom, postulate – a statement taken to be true, though not proven (e.g., Appendix 1)
Lemma – intermediate result proven, then used in proof of a theorem.
Corollary – a theorem that can be established directly as the result of another theorem
Conjecture – statement thought to be true, but not proven to be,
E.g., Collatz Conjecture, twin primes conjecture, Goldbach's conjecture
Definitions
Even integer
n is even if  integer k, s.t., n = 2k.
Odd integer
n is odd if  integer k, s.t., n = 2k+1.
Rational number
real number r is rational if  integers p,q, q ≠ 0, s.t., r = p/q.
Irrational number
real number r is irrational if it is not rational.
FOREWORD: QUANTIFIERS, UI, UG
Quantifiers – yes or no? :: p. 76
– use of UI followed by UG requires choice of arbitrary element.
– this is not the same as choosing any specific element.
Fallacious proof
– To Prove: All integers are even.
– Proof:
UI: Choose any element of the domain, say 44.
Proof step: 44 = 2*22; therefore 44 is even.
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UG: Since true of element chosen, it is true for all elements;
therefore all integers are even!
– Analysis:
The UI/UG combination requires choice of arbitrary element; choosing
a specific element, even if chosen at random, does not satisfy that
requirement.
A correct proof using UI/UG:
Prove: The sum of two odd integers is even.
Proof:
UI:
Let:
a = 2k + 1, where k is any integer
b = 2 l + 1, where l is any integer
Comment: Since we are using the definition of odd integer and k & l are any
arbitrary integers, this sets the stage allowing us to use UG later in the proof.
Then:
Then:
a + b = 2k + 1 + 2
Let:
m=k+
l + 1 = 2(k + l +1)
l +1
Then:
a + b = 2m, which is even by definition.
UG:
For all odd integers, a & b, a + b is even.
Comment: We can apply UG here because a and b were chosen to be
arbitrary odd integers. By using a representation which adheres strictly to the
definition of odd integer, our representation fits all odd integers. And therefore,
anything we derive as true of a and b is also true of all odd integers.
Mathematicians shorthand:
Because the above proof process is used so often, mathematicians usually omit
mentioning all of the things that are obvious by virtue of their frequent use. Thus, the
above proof can be stated much more succinctly, as below.
1. Prove: The sum of two odd integers is even.
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Proof:
Let:
a = 2k + 1
b = 2l + 1
Then:
a + b = 2k + 1 + 2
l + 1 = 2(k + l +1)  EVEN
PROOF METHODS
Direct Proof
Direct proofs start with a proposition (mathematical fact) known to be true and
proceed, step by step, to derive the conclusion directly. The above proof is a good
example of direct proof. It starts with the definition of odd integer and then uses laws
of arithmetic to derive other facts, leading eventually to the desired conclusion.
Indirect Proof – Proof by Contraposition
Proof by contraposition can be used when the proposition to be proved is of the form:
p  q. Recall that the Law of Contraposition states that p  q is equivalent
to q  p. Sometimes the latter is easier to prove than the former. A proof by
contraposition proceeds by showing that if q is true, then p must also be true.
Here is Example #3 from Rosen, p. 78.
Prove: For integer n: 3n + 2 is odd  n is odd.
Prove contrapositive: For integer n: n is even  3n + 2 is even.
Proof:
Let:
n = 2k (definition of even integer)
Then:
3n+2 = 6k+2 = 2(3k+1)
 EVEN
Indirect Proof - Proof by Contradiction
Proof by contradiction is based upon the axiom that every proposition must be either
True or False (Rosen, p. 1). Therefore, if we show that a proposition cannot be false,
then it must be true. Here is Example #10 from Rosen, p. 80.
Prove: 2 is irrational.
Proof by contradiction:
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Assume:
2 is rational.
Then:
There exist integers p & q, q ≠ 0, s.t., 2 = p/q.
WOLG, we assume p & q have no common factors since, any common
factors can be eliminated by expressing p/q in lowest terms.
By the laws of arithmetic:
(2)2 = (p/q)2
2 = p2/q2
2q2 = p2
Thus:
p2 is even
By Exercise 16 (to be done later):
p is even
By definition:
p = 2k, for some integer k.
Therefore:
2q2 = (2k)2= 4k2
q2 = 2k2
And so:
q2 is even; and q is even.
q = 2l for some l.
But we now have a contradiction:
Since p = 2k and q = 2l, they have a common factor, namely 2. 
Therefore:
our assumption that 2 is rational is false; and we have proven that:
2 is irrational.
This proof can also be written in an abbreviated form, as below.
Prove: 2 is irrational.
Proof by contradiction:
Assume:
2 is rational.
Then:
There exist integers p & q, q ≠ 0,
s.t., 2 = p/q, with no common factors for p & q.
By the laws of arithmetic:
(2)2 = (p/q)2
2 = p2/q2
2q2 = p2
Then:
p is even, i.e., p = 2k, for some integer k.
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2q2 = (2k)2= 4k2
q2 = 2k2
Therefore:
q is even, i.e., q = 2 l, for some integer l.
p and q have a common factor, namely 2.
Therefore:
2 is irrational.

Mistakes in proofs:
Division by zero
Example #15, p. 83
Logical fallacies
Affirming the conclusion, Example #16, pp. 83-84
Denying the hypothesis, Example #17, p. 84
Begging the question (circular reasoning), Example #18, p. 84
1.7 Proof Methods and Strategies
PROOF METHODS
Exhaustive Proof
Examine all of a relatively small number of cases; show true to all.
Prove: The only consecutive perfect powers less than 100 are 8 & 9.
{Example 2, p. 87}
Exhaustive proof:
List all powers < 100:
N
N2
N3
N4
1
1
1
1
2
4
8
16
3
9
27
81
4
16
64
5
25
6
36
7
49
8
64
9
81
N5
1
32
N6
1
64
N7
1
Examine this list. The only consecutive integers in that list are 8 & 9.
Proof by Cases:
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Use when no obvious way to prove for all values, but can individual cases readily
provable.
Caveat: Make sure all cases are considered.
Prove: n2 ≥ n, for all integers.
{Example 3, p. 88}
Proof, by cases:
Integers naturally fall into 3 categories: negative, zero, and positive.
Case 1: n < 0.
Then n = -k, for some k ≥ 1.
And n2 = (-k)2 = k2 ≥ 1 > 0 > n.
So n2 ≥ n.
Case 2: n = 0.
Then n2 = 02 = 0 = n.
So n2 ≥ n.
Case 3: n > 0.
Then n ≥ 1.
And n*n ≥ 1*n.
So n2 ≥ n.
Exhaustive Proof Errors
1. Draw a conclusion for some, but not all instances. No matter how many instances
are considered, no conclusion is warranted unless all instances are considered.
Mathematicians sometimes call this the "Law of Small Numbers".
Example 8, p. 90.
2. Attempt proof by cases, but omit some cases.
Example 9, p. 90.
Existence Proofs - Constructive
Provide an element a such that P(a) is true.
Show: There is a positive integer that can be written as the sum of cubes of positive
integers in two different ways.
{Example 10, p. 91}
Proof:
1729 = 103 + 93 = 123 + 13
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Existence Proofs - Nonconstructive
Prove that xP(x) is true. Often this is done via proof by contradiction. I.e., derive a
contradiction from xP(x).
Show: There exist irrational numbers x and y such that xy is rational.
{Example 11, p. 91}
Proof:
We know 2 is irrational.
Consider 22. Either it is rational or irrational.
Case 1: 22 is rational.
Then let x = 2 and y = 2.
Case 2: 22 is irrational.
Then let x = 22 and y = 2.
So xy = (22)2= 22*2= 22= 2.
So either in case 1 or in case 2 we have xy rational, though x and y are
both irrational. Although we do not know which case meets the specification,
we know that one must.
Exercise: Use the calculator to determine which of the two cases meets the
specification.
Uniqueness Proofs
There are two parts to uniqueness proofs:
Existence: Show element with desired property exists.
Uniqueness: Show x ≠ y, x does not have desired property;
or, show if x and y both have desired property, then x = y.
Show: For odd integers a & b, a ≠ b, ! integer c, s.t. |a – c| =|b – c|.
{Exercise 15, p. 103}
Proof:
Existence: We show existence constructively.
WOLG, assume a > b.
If |a – c| = |b – c|, then either a – c = b – c or a – c = c – b.
But a – c = b – c  a = b, so a – c = c – b.
Reworking the equation:
a–c=c–b
a + b = 2c
c = (a + b)/2
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Finally, we know c is an integer since a+b, the sum of two odd integers,
is even.
Uniqueness:
Solutions of linear equations are unique.
Forward and Backward Reasoning:
Start with what you want to prove and working backward derive the premises. If that
succeeds, we construct the proof by reversing the reasoning steps.
Caveat: This only works if at each step we produce statements that are equivalent,
and therefore reversible.
Examples 14 & 15, p. 95 illustrate this process.
Adapting existing proofs:
Often a known proof can be adapted to prove a similar (or parallel) conclusion. This
is illustrated in Example 16, p. 96. Also, in 1.6 homework answers, the proof of
Exercise 1 was adapted in the proofs of Exercises 2 and 6.
Proof strategy in action
On p. 97 Rosen presents many strategies for generating hypotheses and conjectures
and for seeking proofs or counterexamples. Among others, this includes "trying it on
for size," the method applied to Exercise 7 of 1.6. In general, before we attempt a
proof we should look at enough examples to convince us that what we are attempting
to prove is indeed true. Sometime this may surprise us (see immediately below).
Look for counterexamples:
Sometimes, in the process of looking at examples, we will be surprised to find an
example for which the proposed proposition does not hold. This will, of course, be a
counterexample and will enable us to disprove by counterexample. Exercise 38 of
1.6 illustrates this.
Tilings:
Tiling problems can be fiendishly difficult, but sometimes a single observation can
lead to a simple and easy solution. Examples 18-22, pp. 98-100 illustrate.
Role of Open Problems
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There are some fascinating stories in the history of mathematics involving attempts to
solve open problems. Among these are: Fermat's Last Theorem and the Collatz
Conjecture.
One of keen interest to computer scientists is this question: is P = NP, where P is
the set of problems that can be solved in polynomial time and NP is the set of
problems which can be solved in non-deterministic polynomial time. Most computer
scientists and mathematicians think that P ≠ NP, and that problems in the class NP
require exponential (or worse) time. We will look at some exponential and factorial
algorithms later in this course.
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