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1. [(p  q)  q] → p
p
TT
 q   q   p
T
T F
F T
F F
T
F
T
F
T
F
T
F
T
F
F
F
F
T
T
T
F
T
T
The given proposition is a contingent.
2. (p → q)  (q → p)
p
TT
→ q  q  p
T
T F
F T
F F
T
T
F
T
F
T
F
T
F
F
T
T
T
F
F
T
T
F
T
The given proposition is a contingent.
3. [(p → q)  (q → r)] → (p → r)
The given proposition is a tautology.
p → q  q → r → p → r
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
F
F
T
T
T
T
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
F
T
T
T
F
T
T
T
F
F
F
T
F
T
T
T
F
T
F
T
T
T
T
T
T
T
T
T
T
T
T
Two propositions are said to
be logically equivalent iff their
biconditional is a tautology.
We write
A  B
to mean that propositions A
and B are logically equivalent.
5
Illustration: For all propositions p and q,
p  q and ~q  ~p
are logically equivalent.
p
TT
 q  ~ q  ~ p
T
T F
F T
F F
T
T
F
F
T
F
T
T
T
T
T
F
F
T
T
T
F
T
T
6
Illustration: For all propositions p and q,
p  q and q  p
are not logically equivalent.
p
TT
 q  q  p
T
T F
F T
F F
T
T
F
T
F
T
F
T
F
F
T
T
T
F
F
T
T
F
T
7
Rules of Replacement
1. Idempotency (ID)
p  p  p
p  p  p
2. Double Negation (DN)
  p  p
3. De Morgan’s Law (DM)
(p  q)   p  q
(p  q)   p  q
8
4. Material Implication (MI)
p  q   p  q
5. Commutation (CM)
p  q  q  p
p  q  q  p
6. Association (AS)
(p  q)  r  p  (q  r)
(p  q)  r  p  (q  r)
9
7. Distribution (DS)
a. p  (q  r)  (p  q)  (p  r)
b. p  (q  r)  (p  q)  (p  r)
8. Transposition (TR)
p  q   q   p
9. Material Equivalence (ME)
p  q  (p  q)  (q  p)
10
Assignment on
Rules of Replacement
1. Establish Material Implication:
2. Establish Distribution(a)
p  (q  r)  (p  q)  (p  r).
11
p  q   p  q
1.2.4 Deductive arguments
Goal: to investigate how arguments actually
proceed from premises to conclusions. This process
is called inference (we infer the conclusion from
the premises).
Two basic types of inferential processes:
• deductive inference - a specific conclusion is
deduced (or logically derived) from
general premises
• inductive inference - a conclusion is formed by
generalizing from specific premises
12
Example of a deductive argument:
All human beings are mortal.
I am human.
Therefore, I am mortal.
Example of an inductive argument:
Some UP students are war-freak.
Juan is a UP student.
So Juan is war-freak.
13
Deductive arguments with categorical
propositions
Argument 1:
Premise: All UP students are intelligent.
Premise: All intelligent people will succeed in life.
Conclusion: All UP students will succeed in life.
Draw the Venn diagram.
14
Argument 1:
Premise: All UP students are intelligent.
Premise: All intelligent people will succeed in life.
Conclusion: All UP students will succeed in life.
Let U be the set of all UP students,
I be the set of all intelligent people and
S be the set of all people who will succeed in life.
U
I S
15
Question: Is the argument valid?
An argument is valid if its conclusion necessarily
follows from its premises - even though we may not
agree that its premises are true or that its conclusion is
true.
In logic, there is a distinction between validity and
truth. Validity is concerned only with the logical
structure (or form) of an argument, not the truth of its
premises or conclusions.
16
Possibilities for an argument:
1. Valid and sound (logical and with true
premises and conclusion)
2. Valid but not sound (logical but a
premise/conclusion is false)
3. Invalid with a false conclusion
4. Invalid with a true conclusion
17
Argument 2:
Premise: All teen-agers are irresponsible.
Premise: Some girls are teen-agers.
Conclusion: Some girls are irresponsible.
Is this argument valid? I T
G
Yes, it is a valid argument, but is it sound?
18
Argument 3.
Premise: All frogs are mammals.
Premise: All mammals are human beings.
Conclusion: All frogs are human beings.
Is the argument valid? Is it sound?
H
F M
19
Argument 4.
Premise: Mermaids live in the water.
Premise: Mermen are not mermaids.
Conclusion: Mermen do not live in the water.
Is this argument valid? W
G
B
B
B
20
Argument 5. (Argument with a singular proposition).
Premise: All past Philippine Presidents were
from Luzon.
Premise: Erap is from Luzon.
Conclusion: Erap is a former Philippine President.
Is the argument valid? People from Luzon
Phil. Pres.
E
E
21
Deductive Arguments with
one conditional premise
(the four basic conditional
arguments)
22
1. Affirming the antecedent (modus ponens):
Premise: If p, then q.
Premise: p.
Conclusion: q.
This is a valid argument.
23
You can verify that [(p  q )  p]  q
is a tautology.
Example:
Premise: If you pass all your subjects then
your mom will buy you a dog.
Premise: You passed all your subjects.
Conclusion: Your mom will buy you a dog.
2. Affirming the consequent
Premise: If p then q.
Premise: q.
Conclusion: p.
25
This is an invalid argument.
You can verify that [(p  q )  q]  p
is a contingent.
Example:
Premise: If it rains heavily, then the garden
gets wet.
Premise: The garden is wet.
Conclusion: It rained heavily.
This is an invalid argument. The conclusion
does not follow from the premises.
3. Denying the consequent (modus tollens)
Premise: If p then q.
Premise: Not q.
Conclusion: Not p
This is a valid argument.
Example:
If today is Wednesday, then tomorrow is Thursday.
Tomorrow is not Thursday.
Therefore, today is not Wednesday.
4. Denying the antecedent
Premise: If p then q.
Premise: Not p .
Conclusion: Not q.
This is not valid.
Example:
If you like the book, then you’ll love the movie.
You did not like the book.
Therefore, you will not love the movie.
28
Deductive Arguments with a chain of
conditionals
If p then q.
If q then r.
If p then r.
Example:
“If there is a typhoon in the North, then there will
be a shortage of cabbages.
A shortage of cabbages means high prices for
cabbages.
Therefore, a typhoon in the north means high
prices for cabbages.”
29
Example:
“If a function is differentiable at a number,
then the function is continuous at that number.
If a function is continuous at a number,
then the limit of the function as x approaches that
number exists.
Therefore, differentiability at a number
implies the existence of the limit of the function
at that number.
30
31
1.2.5 Inductive Arguments
Consider the following example of an inductive
argument.
Birds fly up into the air but
eventually come back down.
People or animals that jump or are thrown into the air
fall back down.
Rocks thrown into the air
come back down.
Balls thrown into the air come back down.
So what goes up must come down.
Note: Each premise represents a specific case or example of
something that goes up and which comes back down. The
conclusion represents a generalization of these specific cases.
Premise: Birds fly up into the air but eventually
come back down.
Premise: People or animals that jump into the air
fall back down.
Premise: Rocks thrown into the air come back down.
Premise: Balls thrown into the air come back down.
Conclusion: Everything that goes up must come down.
A Comparison of Deductive and Inductive Arguments
Deductive
• The conclusion usually is
more specific than the
premises.
Inductive
• The conclusion usually is
more general than the
premises.
• In a valid deductive
argument, the conclusion
necessarily follows from
the premises.
• There is no such thing as a valid
inductive argument. Inductive
arguments can be analyzed only in
terms of their strength, that is, we
make a subjective judgment about
how well the premises support the
generalization in the conclusion.
The conclusion of a strong
inductive argument seems likely to
follow from its premises, but it
does not necessarily do so.
35
Induction and Deduction in Everyday Life
People usually form reasoned opinions and
decisions through inductive reasoning. It helps a
person to organize knowledge and suggest
possible truths.
But proof requires deduction, in which a
conclusion is necessarily established from a set
of premises. Deduction allows a person to prove
possible truths.
36
Deduction and Induction in Mathematics
Theorems are statements of mathematical truth which
requires proof which is possible only through deductive
logic.
Axioms are the starting points for mathematical proof, the
“givens”, assumed to be true without proof.
Although each proof is deductive, induction also plays a role:
ideas for theorems usually come through inductive reasoning.
Example: Goldbach Conjecture (1742)
“Every even number (except 2) can be expressed as a sum
of two prime numbers.”
37
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7
12 = 5 + 7
14 = 7 + 7
38
Goldbach’s conjecture
remains an open problem:
No proof nor
counterexample is found
yet.
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
2
21

2
32

2
43

2
54

2
65

 n...21
Conjecture
 
2
1 nn
Rules of Inference
1. Conjunction (CJ)
p
q
 p  q
Illustration:
2 is even.
2 is prime
 2 is even and 2 is prime.
or
 2 is both prime and even.
2. Simplification (SP)
p  q
 p
Illustration:
2 is even and 2 is prime.
 2 is even.
 2 is prime
p  q
 q
p
 p  q
Illustration:
2 is even.
 2 is even or 3 is even.
3. Addition (AD)
p → q
p
 q
Illustration:
If a number is even, then its square is even.
2 is even.
 The square of 2 is even.
4. Modus Ponens (MP)
p → q
q
 p
Illustration:
If a number is even, then its square is even.
The square of 3 is not even.
 3 is not even.
5. Modus Tollens (MT)
p  q
p
 q
Illustration:
A function is either algebraic or transcendental.
An exponential function is not algebraic.
 An exponential function is transcendental.
6. Disjunctive Syllogism (DS)
p → q
q → r
 p → r
Illustration:
Differentiability implies continuity.
Continuity implies integrability.
 Differentiability implies integrability.
7. Hypothetical Syllogism (HS)
Assignment No. 3
Rules of Inference
1. Establish Disjunctive Syllogism (DS)
p  q
p
 q
(Show that [(p  q)  p]→ q is a tautology.)
2. Establish Hypothetical Syllogism (HS)
p → q
q → r
 p → r
(Show that [(p → q)  (q →r)]→ (p →r)
is a tautology.)
Mathematical theorems are either
1. conditional statements (If p then q)
or
2. biconditional statements (p iff q).
We will study how to prove a conditional
statement.
The proof of a biconditional statement
p iff q
may consist of the proofs of 2 conditional
statements
“If p then q” and “If q then p”.
(p  q)  (p  q)  (q  p)
Theorem: If p then q.
Methods of proof:
1. Direct 2. Indirect
(prf by contradiction)
Proof:
Suppose p.
.
.
.
q.
Proof:
Suppose p q.
.
.
c  c
q.
3. Transposition
Proof:
Suppose q.
.
.
.
p.
51
Theorem. If an integer x is even,
then x2 is even.
Proof: (direct method)
Suppose x is even integer.
By definition, x = 2n for some integer n.
Thus, x2 = (2n)2 .
So x2 = (2n)2 = 4n2 = 2(2n2).
Since n is an integer, 2n2 is also an integer.
Since x2 is twice an integer, x2 is also even.
QED 52
Proof:
Suppose p.
.
.
.
q.
Let a and b be real numbers
and c be a non-zero real number. Then
Theorem: If a and b are real numbers
and c is a non-zero real number, then
.
c
b
c
a
c
ba


proof: (direct method)
 
c
ba
c
ba 1


(by definition of division)
c
b
c
a
11
 (by RHDPMA)
.
c
b
c
a
 (by definition of division)
53
Proof:
Suppose p.
.
.
.
q.
     .CABACBA 
For all sets A, B and C,
Restatement:
     .CABACBA 
If A, B and C are sets, then
proof: (direct method)
Let A, B and C be sets. Then
 CBA 
  CBAx:Ux 
  CBxAxUx  and:
  CxBxAxUx  orand:
    CxAxBxAxUx  andorand:
    CAxBAxUx  or:
    CABAxUx  :
   .CABA 
(by definition of )
(by definition of )
(by definition of )
(by DT)
(by definition of )
(by definition of )
(by definition of )
For any   k where k is any integer,
.
sin
cos
cos
sin



 


1
1
Wrong proof:
.
sin
cos
cos
sin



 


1
1
  θcosθcosθsin  11
?
2
θcosθsin 2
?
2
1
θsinθsin 22


For any   k where k is any integer,
.
sin
cos
cos
sin



 


1
1
proof: (direct method)
Suppose   k where k is any integer.
Then
(-1,0) (1,0)

1cos
.sin 0
and
 01 cos

 

cos
sin
1 



cos
cos
cos
sin



 1
1
1
 


2
1
1
cos
cossin



 


2
1
sin
cossin 



sin
cos

1
,cos 01Since  .
cos
cos
1
1
1





Theorem. If a2 is even, then a is even.
Proof: (indirect method)
Suppose a2 is even and a is odd.
Since a is odd, then by definition, we can find an
integer k such that
a = 2k + 1
 a2 = (2k + 1)2
= 4k2 + 4k+1
= 2(2k2 + 2k) +1
By definition, a2 is odd.
59
Proof:
Suppose p q.
.
.
c  c
q.
So a2 is even and a2 is odd, a contradiction.
QED
So (p  q) is true.
We wanted to prove that p  q.
We started with p  q.
But we arrived at a contradiction.
It means that p  q is false.
(p  q)
 p  q
  p    q
  p  q
60
(De Morgan’s Law)
(Double Negation)
(Material Implication)
Theorem. .irrationalis2
Theorem. If we assume everything that is
true about numbers, then
.irrationalis2
Proof: (indirect method)
Suppose what we know about numbers is
true and is rational.2
By definition, there exist integers p and q
such that
.2
q
p
Without loss of generality, we may further
assume that p and q are relatively prime, that is
they have no common factor except 1.
,2Since 
q
p
 2
2
2





q
p
2
2
2

q
p 22
2qp 
,22
2Since qp  even.is2
p
even.alsoistheneven,isSince 2
pp
even,isSince p such thatintegeranexiststhere k
.2kp 
Then
  22
22 qk  22
24 qk  .22
2 qk 
,22
2Since kq  even.is2
q
even.alsoistheneven,isSince 2
qq
even.bothareandThus, qp
ion.contradictaeven,bothare
andprimerelativelyareandThen qp
.irrationalis2Thus,
QED
Theorem. If a2 is even, then a is even.
Proof: (transposition method)
Suppose a is odd.
Since a is odd, then by definition, we can find an
integer k such that
a = 2k + 1
 a2 = (2k + 1)2
= 4k2 + 4k+1
= 2(2k2 + 2k) +1
By definition, a2 is odd.
65
QED
Proof:
Suppose q.
.
.
.
p.
Usefulness of seeking inductive
evidence:
A mathematical rule can be
tested inductively. Although test cases
constitute inductive evidence only,
and not proof, they often are enough
to satisfy yourself of a rule’s truth.
.a
a


1
3
3
67
A proposed rule can be invalidated by one
failed test case.
Conjecture: For any real number a,
The conjecture is false since if a = 1, we
get 4/3 = 2.
Fermat’s Last Theorem
(Pierre Fermat, 1601-1665)
“For any natural number n besides 1 or 2, it
is impossible to find natural numbers a, b,
and c that satisfy the relationship
an + bn = cn.”
68
• This theorem was first conjectured by Pierre de
Fermat in 1637, famously in the margin of a
copy of Arithmetica where he claimed he had a
proof that was too large to fit in the margin.
• No successful proof was published until 1995 despite
the efforts of many mathematicians. The
unsolved problem stimulated the development of
algebraic number theory in the 19th century and
the proof of the modularity theorem in the 20th.
• It is among the most famous theorems in the history
of mathematics and prior to its 1995 proof was
in the Guinness book of World Records for
"most difficult math problem".
In 1993, Andrew Wiles
presented his proof to the public
for the first time at a conference in
Cambridge.
In August 1993, however, it
turned out that the proof contained
a gap.
Together with his former
student Richard Taylor, he
published a second paper which
circumvented the gap and thus
completed the proof.
Both papers were published in
1995 in a special volume of the
Annals of Mathematics.
Andrew Wiles

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chapter 1 (part 2)

  • 1.
  • 2. 1. [(p  q)  q] → p p TT  q   q   p T T F F T F F T F T F T F T F T F F F F T T T F T T The given proposition is a contingent.
  • 3. 2. (p → q)  (q → p) p TT → q  q  p T T F F T F F T T F T F T F T F F T T T F F T T F T The given proposition is a contingent.
  • 4. 3. [(p → q)  (q → r)] → (p → r) The given proposition is a tautology. p → q  q → r → p → r T T T T F F F F T T F F T T F F T F T F T F T F T T F F T T T T T T T T F F F F T T F F T T F F T F T F T F T F T F T T T F T T T F F F T F T T T F T F T T T T T T T T T T T T
  • 5. Two propositions are said to be logically equivalent iff their biconditional is a tautology. We write A  B to mean that propositions A and B are logically equivalent. 5
  • 6. Illustration: For all propositions p and q, p  q and ~q  ~p are logically equivalent. p TT  q  ~ q  ~ p T T F F T F F T T F F T F T T T T T F F T T T F T T 6
  • 7. Illustration: For all propositions p and q, p  q and q  p are not logically equivalent. p TT  q  q  p T T F F T F F T T F T F T F T F F T T T F F T T F T 7
  • 8. Rules of Replacement 1. Idempotency (ID) p  p  p p  p  p 2. Double Negation (DN)   p  p 3. De Morgan’s Law (DM) (p  q)   p  q (p  q)   p  q 8
  • 9. 4. Material Implication (MI) p  q   p  q 5. Commutation (CM) p  q  q  p p  q  q  p 6. Association (AS) (p  q)  r  p  (q  r) (p  q)  r  p  (q  r) 9
  • 10. 7. Distribution (DS) a. p  (q  r)  (p  q)  (p  r) b. p  (q  r)  (p  q)  (p  r) 8. Transposition (TR) p  q   q   p 9. Material Equivalence (ME) p  q  (p  q)  (q  p) 10
  • 11. Assignment on Rules of Replacement 1. Establish Material Implication: 2. Establish Distribution(a) p  (q  r)  (p  q)  (p  r). 11 p  q   p  q
  • 12. 1.2.4 Deductive arguments Goal: to investigate how arguments actually proceed from premises to conclusions. This process is called inference (we infer the conclusion from the premises). Two basic types of inferential processes: • deductive inference - a specific conclusion is deduced (or logically derived) from general premises • inductive inference - a conclusion is formed by generalizing from specific premises 12
  • 13. Example of a deductive argument: All human beings are mortal. I am human. Therefore, I am mortal. Example of an inductive argument: Some UP students are war-freak. Juan is a UP student. So Juan is war-freak. 13
  • 14. Deductive arguments with categorical propositions Argument 1: Premise: All UP students are intelligent. Premise: All intelligent people will succeed in life. Conclusion: All UP students will succeed in life. Draw the Venn diagram. 14
  • 15. Argument 1: Premise: All UP students are intelligent. Premise: All intelligent people will succeed in life. Conclusion: All UP students will succeed in life. Let U be the set of all UP students, I be the set of all intelligent people and S be the set of all people who will succeed in life. U I S 15
  • 16. Question: Is the argument valid? An argument is valid if its conclusion necessarily follows from its premises - even though we may not agree that its premises are true or that its conclusion is true. In logic, there is a distinction between validity and truth. Validity is concerned only with the logical structure (or form) of an argument, not the truth of its premises or conclusions. 16
  • 17. Possibilities for an argument: 1. Valid and sound (logical and with true premises and conclusion) 2. Valid but not sound (logical but a premise/conclusion is false) 3. Invalid with a false conclusion 4. Invalid with a true conclusion 17
  • 18. Argument 2: Premise: All teen-agers are irresponsible. Premise: Some girls are teen-agers. Conclusion: Some girls are irresponsible. Is this argument valid? I T G Yes, it is a valid argument, but is it sound? 18
  • 19. Argument 3. Premise: All frogs are mammals. Premise: All mammals are human beings. Conclusion: All frogs are human beings. Is the argument valid? Is it sound? H F M 19
  • 20. Argument 4. Premise: Mermaids live in the water. Premise: Mermen are not mermaids. Conclusion: Mermen do not live in the water. Is this argument valid? W G B B B 20
  • 21. Argument 5. (Argument with a singular proposition). Premise: All past Philippine Presidents were from Luzon. Premise: Erap is from Luzon. Conclusion: Erap is a former Philippine President. Is the argument valid? People from Luzon Phil. Pres. E E 21
  • 22. Deductive Arguments with one conditional premise (the four basic conditional arguments) 22
  • 23. 1. Affirming the antecedent (modus ponens): Premise: If p, then q. Premise: p. Conclusion: q. This is a valid argument. 23 You can verify that [(p  q )  p]  q is a tautology.
  • 24. Example: Premise: If you pass all your subjects then your mom will buy you a dog. Premise: You passed all your subjects. Conclusion: Your mom will buy you a dog.
  • 25. 2. Affirming the consequent Premise: If p then q. Premise: q. Conclusion: p. 25 This is an invalid argument. You can verify that [(p  q )  q]  p is a contingent.
  • 26. Example: Premise: If it rains heavily, then the garden gets wet. Premise: The garden is wet. Conclusion: It rained heavily. This is an invalid argument. The conclusion does not follow from the premises.
  • 27. 3. Denying the consequent (modus tollens) Premise: If p then q. Premise: Not q. Conclusion: Not p This is a valid argument. Example: If today is Wednesday, then tomorrow is Thursday. Tomorrow is not Thursday. Therefore, today is not Wednesday.
  • 28. 4. Denying the antecedent Premise: If p then q. Premise: Not p . Conclusion: Not q. This is not valid. Example: If you like the book, then you’ll love the movie. You did not like the book. Therefore, you will not love the movie. 28
  • 29. Deductive Arguments with a chain of conditionals If p then q. If q then r. If p then r. Example: “If there is a typhoon in the North, then there will be a shortage of cabbages. A shortage of cabbages means high prices for cabbages. Therefore, a typhoon in the north means high prices for cabbages.” 29
  • 30. Example: “If a function is differentiable at a number, then the function is continuous at that number. If a function is continuous at a number, then the limit of the function as x approaches that number exists. Therefore, differentiability at a number implies the existence of the limit of the function at that number. 30
  • 31. 31 1.2.5 Inductive Arguments Consider the following example of an inductive argument. Birds fly up into the air but eventually come back down.
  • 32. People or animals that jump or are thrown into the air fall back down. Rocks thrown into the air come back down.
  • 33. Balls thrown into the air come back down. So what goes up must come down.
  • 34. Note: Each premise represents a specific case or example of something that goes up and which comes back down. The conclusion represents a generalization of these specific cases. Premise: Birds fly up into the air but eventually come back down. Premise: People or animals that jump into the air fall back down. Premise: Rocks thrown into the air come back down. Premise: Balls thrown into the air come back down. Conclusion: Everything that goes up must come down.
  • 35. A Comparison of Deductive and Inductive Arguments Deductive • The conclusion usually is more specific than the premises. Inductive • The conclusion usually is more general than the premises. • In a valid deductive argument, the conclusion necessarily follows from the premises. • There is no such thing as a valid inductive argument. Inductive arguments can be analyzed only in terms of their strength, that is, we make a subjective judgment about how well the premises support the generalization in the conclusion. The conclusion of a strong inductive argument seems likely to follow from its premises, but it does not necessarily do so. 35
  • 36. Induction and Deduction in Everyday Life People usually form reasoned opinions and decisions through inductive reasoning. It helps a person to organize knowledge and suggest possible truths. But proof requires deduction, in which a conclusion is necessarily established from a set of premises. Deduction allows a person to prove possible truths. 36
  • 37. Deduction and Induction in Mathematics Theorems are statements of mathematical truth which requires proof which is possible only through deductive logic. Axioms are the starting points for mathematical proof, the “givens”, assumed to be true without proof. Although each proof is deductive, induction also plays a role: ideas for theorems usually come through inductive reasoning. Example: Goldbach Conjecture (1742) “Every even number (except 2) can be expressed as a sum of two prime numbers.” 37
  • 38. 4 = 2 + 2 6 = 3 + 3 8 = 3 + 5 10 = 3 + 7 12 = 5 + 7 14 = 7 + 7 38 Goldbach’s conjecture remains an open problem: No proof nor counterexample is found yet.
  • 39. 1 = 1 1 + 2 = 3 1 + 2 + 3 = 6 1 + 2 + 3 + 4 = 10 1 + 2 + 3 + 4 + 5 = 15 2 21  2 32  2 43  2 54  2 65   n...21 Conjecture   2 1 nn
  • 40. Rules of Inference 1. Conjunction (CJ) p q  p  q Illustration: 2 is even. 2 is prime  2 is even and 2 is prime. or  2 is both prime and even.
  • 41. 2. Simplification (SP) p  q  p Illustration: 2 is even and 2 is prime.  2 is even.  2 is prime p  q  q
  • 42. p  p  q Illustration: 2 is even.  2 is even or 3 is even. 3. Addition (AD)
  • 43. p → q p  q Illustration: If a number is even, then its square is even. 2 is even.  The square of 2 is even. 4. Modus Ponens (MP)
  • 44. p → q q  p Illustration: If a number is even, then its square is even. The square of 3 is not even.  3 is not even. 5. Modus Tollens (MT)
  • 45. p  q p  q Illustration: A function is either algebraic or transcendental. An exponential function is not algebraic.  An exponential function is transcendental. 6. Disjunctive Syllogism (DS)
  • 46. p → q q → r  p → r Illustration: Differentiability implies continuity. Continuity implies integrability.  Differentiability implies integrability. 7. Hypothetical Syllogism (HS)
  • 47. Assignment No. 3 Rules of Inference 1. Establish Disjunctive Syllogism (DS) p  q p  q (Show that [(p  q)  p]→ q is a tautology.)
  • 48. 2. Establish Hypothetical Syllogism (HS) p → q q → r  p → r (Show that [(p → q)  (q →r)]→ (p →r) is a tautology.)
  • 49. Mathematical theorems are either 1. conditional statements (If p then q) or 2. biconditional statements (p iff q). We will study how to prove a conditional statement.
  • 50. The proof of a biconditional statement p iff q may consist of the proofs of 2 conditional statements “If p then q” and “If q then p”. (p  q)  (p  q)  (q  p)
  • 51. Theorem: If p then q. Methods of proof: 1. Direct 2. Indirect (prf by contradiction) Proof: Suppose p. . . . q. Proof: Suppose p q. . . c  c q. 3. Transposition Proof: Suppose q. . . . p. 51
  • 52. Theorem. If an integer x is even, then x2 is even. Proof: (direct method) Suppose x is even integer. By definition, x = 2n for some integer n. Thus, x2 = (2n)2 . So x2 = (2n)2 = 4n2 = 2(2n2). Since n is an integer, 2n2 is also an integer. Since x2 is twice an integer, x2 is also even. QED 52 Proof: Suppose p. . . . q.
  • 53. Let a and b be real numbers and c be a non-zero real number. Then Theorem: If a and b are real numbers and c is a non-zero real number, then . c b c a c ba   proof: (direct method)   c ba c ba 1   (by definition of division) c b c a 11  (by RHDPMA) . c b c a  (by definition of division) 53 Proof: Suppose p. . . . q.
  • 54.      .CABACBA  For all sets A, B and C, Restatement:      .CABACBA  If A, B and C are sets, then proof: (direct method) Let A, B and C be sets. Then
  • 55.  CBA    CBAx:Ux    CBxAxUx  and:   CxBxAxUx  orand:     CxAxBxAxUx  andorand:     CAxBAxUx  or:     CABAxUx  :    .CABA  (by definition of ) (by definition of ) (by definition of ) (by DT) (by definition of ) (by definition of ) (by definition of )
  • 56. For any   k where k is any integer, . sin cos cos sin        1 1 Wrong proof: . sin cos cos sin        1 1   θcosθcosθsin  11 ? 2 θcosθsin 2 ? 2 1 θsinθsin 22  
  • 57. For any   k where k is any integer, . sin cos cos sin        1 1 proof: (direct method) Suppose   k where k is any integer. Then (-1,0) (1,0)  1cos .sin 0 and  01 cos
  • 58.     cos sin 1     cos cos cos sin     1 1 1     2 1 1 cos cossin        2 1 sin cossin     sin cos  1 ,cos 01Since  . cos cos 1 1 1     
  • 59. Theorem. If a2 is even, then a is even. Proof: (indirect method) Suppose a2 is even and a is odd. Since a is odd, then by definition, we can find an integer k such that a = 2k + 1  a2 = (2k + 1)2 = 4k2 + 4k+1 = 2(2k2 + 2k) +1 By definition, a2 is odd. 59 Proof: Suppose p q. . . c  c q. So a2 is even and a2 is odd, a contradiction. QED
  • 60. So (p  q) is true. We wanted to prove that p  q. We started with p  q. But we arrived at a contradiction. It means that p  q is false. (p  q)  p  q   p    q   p  q 60 (De Morgan’s Law) (Double Negation) (Material Implication)
  • 61. Theorem. .irrationalis2 Theorem. If we assume everything that is true about numbers, then .irrationalis2 Proof: (indirect method) Suppose what we know about numbers is true and is rational.2 By definition, there exist integers p and q such that
  • 62. .2 q p Without loss of generality, we may further assume that p and q are relatively prime, that is they have no common factor except 1. ,2Since  q p  2 2 2      q p 2 2 2  q p 22 2qp 
  • 63. ,22 2Since qp  even.is2 p even.alsoistheneven,isSince 2 pp even,isSince p such thatintegeranexiststhere k .2kp  Then   22 22 qk  22 24 qk  .22 2 qk  ,22 2Since kq  even.is2 q even.alsoistheneven,isSince 2 qq
  • 65. Theorem. If a2 is even, then a is even. Proof: (transposition method) Suppose a is odd. Since a is odd, then by definition, we can find an integer k such that a = 2k + 1  a2 = (2k + 1)2 = 4k2 + 4k+1 = 2(2k2 + 2k) +1 By definition, a2 is odd. 65 QED Proof: Suppose q. . . . p.
  • 66. Usefulness of seeking inductive evidence: A mathematical rule can be tested inductively. Although test cases constitute inductive evidence only, and not proof, they often are enough to satisfy yourself of a rule’s truth.
  • 67. .a a   1 3 3 67 A proposed rule can be invalidated by one failed test case. Conjecture: For any real number a, The conjecture is false since if a = 1, we get 4/3 = 2.
  • 68. Fermat’s Last Theorem (Pierre Fermat, 1601-1665) “For any natural number n besides 1 or 2, it is impossible to find natural numbers a, b, and c that satisfy the relationship an + bn = cn.” 68
  • 69. • This theorem was first conjectured by Pierre de Fermat in 1637, famously in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. • No successful proof was published until 1995 despite the efforts of many mathematicians. The unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th. • It is among the most famous theorems in the history of mathematics and prior to its 1995 proof was in the Guinness book of World Records for "most difficult math problem".
  • 70. In 1993, Andrew Wiles presented his proof to the public for the first time at a conference in Cambridge. In August 1993, however, it turned out that the proof contained a gap. Together with his former student Richard Taylor, he published a second paper which circumvented the gap and thus completed the proof. Both papers were published in 1995 in a special volume of the Annals of Mathematics. Andrew Wiles