Chapter 1. The logic and language of proofs
Foundation of Mathematics
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
Chapter 1. The logic and language of proofs
§1.1 Propositions
§1.2 Expressions and tautologies
§1.3 Quantifies
§1.4 Methods of proofs
§1.5 The contradiction method of proof
§1.6 More proofs
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.1 Propositions
Definition
A proposition (or statement) is a sentence that is either true or false.
Examples:
“1+2=4”: a proposition
“E. coli is an animal.”: a proposition
“the north Pacific right whale will be extinct species before year 2525.”: a
proposition. Although it takes time to determine the truth value.
“Euclid was left-handed.”: a proposition. It has one truth value but may
never be known.
“x
¡ 2”: not a proposition. 7
x can be anything.
“My calculus teacher is awesome.”: not a proposition.
(a) we do not know who are you? (b) One may agree while one may not.
“What time is it?”: not a proposition. No question can be a proposition.
“Jack is lying about he is a liar.”: not a proposition.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.1 Propositions
We will use letters P, Q, R etc., to denote propositions.
Definition
A negation of a proposition P, denoted by
P, means “not P”.
Examples:
Let P: “Money grows on trees.”
6 P: “Money does not grow on trees.”
Let Q: “2 is an integer.”
6 Q: “2 is not an integer.”
The true table of negation:
P
T
F
Bo-Chih Huang
P
F
T
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.1 Propositions
Definition
Conjunction of P and Q, denoted by P
^ Q, means “P and Q”.
Remark:
^ Q is true if and only if P is true and Q is true.
P ^ Q is false if and only if at least one of P or Q is false.
P
Examples:
Let P : “2019.02.21 is Thursday.” Q: “5 4 ¡ 3.”
6 P ^ Q: “2019.02.21 is Thursday and 5 4 ¡ 3.” So P
The true table of conjunction:
P
T
T
F
F
Bo-Chih Huang
Q
T
F
T
F
P
^Q
T
F
F
F
Foundation of Mathematics
^ Q is true.
Chapter 1. The logic and language of proofs
§1.1 Propositions
Definition
Disjunction of P and Q, denoted by P
_ Q, means “P or Q”.
Remark:
_ Q is true if and only if at least one of P or Q is true
P _ Q is false if and only if both P and Q are false.
P
Examples:
Let P : “2019.02.21 is Monday.” Q: “5 4 ¡ 3.”
6 P _ Q: “2019.02.21 is Monday or 5 4 ¡ 3.” So P
The true table of disjunction:
P
T
T
F
F
Bo-Chih Huang
Q
T
F
T
F
P
_Q
T
T
T
F
Foundation of Mathematics
_ Q is true.
Chapter 1. The logic and language of proofs
§1.1 Propositions
Definition
For propositions P and Q, the conditional sentence P Ñ Q is the proposition
“If P, then Q.” (or “P implies Q.”, “P only if Q.”). Proposition P is called
the hypothesis, and Q is called the conclusion. Moreover, the converse of
P Ñ Q is Q Ñ P, and the contrapositive of P Ñ Q is Q Ñ P.
Examples:
P: “Today is Thursday.”; Q: “I go back to Taipei to play badminton after
calculus class.” Thus, P Ñ Q: “If today is Thursday, then I go back to
Taipei to play badminton after calculus class.”
Ñ
(a) If today is Thursday and I go back to Taipei to play badminton after
calculus class, then P
Q is true. If today is Thursday but I stay in Chiayi
after calculus class, then P
Q is false.
(b) If today is not Thursday, the sentence P
Q state only that something
would happen on Thursday. Whether I am in Taipei or Chiayi (not in
Taipei) is irrelevant. So P
Q is still true.
Ñ
Ñ
Ñ
The true table of implication:
P
T
T
F
F
Bo-Chih Huang
Q
T
F
T
F
P
ÑQ
T
F
T
T
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.1 Propositions & §1.2 Expressions and tautologies
Definition
For propositions P and Q, the biconditional sentence P Ø Q is the proposition
“P if and only if Q.” The sentence P Ø Q is true exactly when P and Q have
the same truth values.
Definition
If we regard P, Q, R as variables, then a combination of variables and
operations, for instance P Ñ Q, P _ pQ ^ R q, , is called a propositional
expression. Two expressions X and Y are logically equivalent or simply
equivalent, denoted by X ô Y , provided they have the same true values for all
possible values of true or false for all variables appearing in either expression.
Example: pP Ø Q q ôrpP Ñ Q q ^ pQ Ñ P qs.
P Q P Ñ Q Q Ñ P P Ø Q pP Ñ Q q ^ pQ
T T
T
T
T
T
T
F
F
T
F
F
F
T
T
F
F
F
F
F
T
T
T
T
Bo-Chih Huang
Foundation of Mathematics
Ñ Pq
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Example:
p
P
P
T
T
F
F
P
_ Q q ôpP Ñ Q q.
Q
P
P _Q
T
F
T
F
Ñ Q and
P
T
T
F
F
Q
T
F
T
F
F
F
T
T
P
P
F
F
T
T
T
F
T
T
Ñ
P
ÑQ
T
F
T
T
Q are not equivalent.
Q PÑQ
PÑ Q
F
T
T
T
F
T
F
T
F
T
T
T
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Definition
A propositional expression is call a tautology if it yields a true propositional
regardless of what propositions replaced its variables. A propositional
expression is called a contradiction if it yields a false proposition regardless of
what propositions replace its variables.
Examples:
P
P
_
^
P
T
T
F
F
P is a tautology
P is a contradiction
P P_ P P^ P
F
T
F
F
T
F
T
T
F
T
T
F
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
P
Ñ pP _ Q q (or Q Ñ pP _ Q q)
P Q P _ Q P Ñ pP _ Q q
Q
Ñ pP _ Q q
T
T
F
F
T
F
T
F
T
T
T
F
T
T
T
T
T
T
T
T
T
T
F
F
T
F
T
F
T
F
F
F
T
T
T
T
T
T
T
T
P
T
T
F
F
Q
T
F
T
F
P
F
F
T
T
pP ^ Q q Ñ P (or pP ^ Q q Ñ Q)
P Q P ^ Q pP ^ Q q Ñ P pP ^ Q q Ñ Q
pP ^ pP Ñ Q qq Ñ Q
Q
F
T
F
T
P
ÑQ
T
F
T
T
Bo-Chih Huang
Q
Ñ
P
pP Ñ Q q Øp
T
F
T
T
Foundation of Mathematics
T
T
T
T
Q
Ñ
Pq
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Proposition 1.1
Each of the following propositional expressions is a tautology
(a) pP _ Q q Øp P Ñ Q q.
(b) pP Ñ Q q ØpP ^ Q q.
+
(e)
pP _ Q q Øp P ^ Q q.
pDe Morgan’s Lawq
pP ^ Q q Øp P _ Q q.
p P q Ø P. pDouble Negation Lawq
Proof
P
T
T
F
F
of (c):
Q
P
T
F
F
F
T
T
F
T
(c)
(d)
l
Q
F
T
F
T
P
_Q
pP _ Q q
T
T
T
F
F
F
F
T
P
^
Q
pP _ Q q Øp
F
F
F
T
The proof of the remaining parts of Proposition 1.1 as an exercise.
Bo-Chih Huang
Foundation of Mathematics
T
T
T
T
P
^
Qq
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Remark
Proposition 1.1(b) tells us how to deny an “if, then” statement. Consider the
statement “If today is Thursday, then I go back to Taipei to play badminton
after calculus class.”, which is of the form P Ñ Q, where P is the statement
“Today is Thursday.” and Q is “I go back to Taipei to play badminton after
calculus class.”. According to Proposition 1.1(b), the denial is P ^ Q; that is,
“Today is Thursday, and I don’t go back to Taipei to play badminton after
calculus class.”
P
T
T
F
F
Q
T
F
T
F
P
F
F
T
T
Q
F
T
F
T
pP Ñ Q q
F
T
F
F
Bo-Chih Huang
P
ÑQ
T
T
T
F
P
Ñ
T
T
F
T
Foundation of Mathematics
Q
P
Ñ
F
T
T
T
Q
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Proposition 1.2
Each of the following propositional expressions is a tautology
(a) P Ñ P.
(b) pP Ø Q q Ñ pQ Ø P q.
(c) pP Ø Q q Øp P Ø Q q.
(d) rpP Ñ Q q ^ pQ Ñ P qs ØpP Ø Q q.
+
rpP Ñ Q q ^ pQ Ñ R qs Ñ pP Ñ R q.
pTransitive Lawq
rpP Ø Q q ^ pQ Ø R qs Ñ pP Ø R q.
+
(g) rP _ pQ ^ R qs ØrpP _ Q q ^ pP _ R qs.
pDistributive Lawq
(h) rP ^ pQ _ R qs ØrpP ^ Q q _ pP ^ R qs.
+
(i) rpP _ Q q _ R s ØrP _ pQ _ R qs.
pAssociative Lawq
(j) rpP ^ Q q ^ R s ØrP ^ pQ ^ R qs.
(k) tp P q Ñ rQ _ p Q qsu Ñ P.
(e)
(f)
The proof of Proposition 1.2 is left as an exercise.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Proposition 1.3
Each of the following propositional expressions is a tautology
(a) pP Ø Q q Ñ pR ^ P Ø R ^ Q q.
(b) pP Ø Q q Ñ pR _ P Ø R _ Q q.
The proof of Proposition 1.3 is left as an exercise.
Proposition 1.4
Each of the following propositional expressions is a contradiction
(a) pP Ñ Q q ^ pP ^ Q q.
(b) rpP _ Q q ^ P s ^ p Q q.
(c) pP ^ Q q ^ p P q.
Proof of (a):
By Proposition 1.1(b), pP ^ Q q Ø pP Ñ Q q. By definition,
pP Ñ Q q ^ pP Ñ Q q is a contradiction. Therefore, pP Ñ Q q ^ pP
a contradiction by Proposition 1.3(a).
Bo-Chih Huang
Foundation of Mathematics
^
Q q is
l
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Let us explain how the proof of Proposition 1.4(a) work.
By definition, for any proposition X , we have the form
pP Ñ Q q ^
pP Ñ Q q
is a contradiction.
By Proposition 1.3(a) for any propositions X , Y , Z , we have the form
rpP ^
Qq Ø
pP Ñ Q qs Ñ rpP Ñ Q q ^ pP ^
is a tautology
Since P ^
statement
QØ
Since pP Ñ Q q ^
is a contradiction.
Ñ Q q ^ pP Ñ Q qs
pP Ñ Q q is a tautology by Proposition 1.1(b), so the
rpP Ñ Q q ^ pP ^
must be true.
Q q Ø pP
Q qs ØrpP
Ñ Q q ^ pP Ñ Q qs
pP Ñ Q q is a contradiction, so pP Ñ Q q ^ pP ^
Bo-Chih Huang
Foundation of Mathematics
Qq
Chapter 1. The logic and language of proofs
§1.2 Expressions and tautologies
Remark
We also can construct the truth table for pP Ñ Q q ^ pP ^ Q q as
P Q
Q P Ñ Q P ^ Q pP Ñ Q q ^ pP ^ Q q
F
T T
F
T
F
T
F
T
F
T
F
F
T
F
T
F
F
F
F
T
T
F
F
Alternately, we can prove Proposition 1.1(c) from Propositions 1.1(a),
1.2(c), 1.1(b), and 1.2(f) as following:
By Proposition 1.1(a), P _ Q Ø P Ñ Q. Therefore, by Proposition
1.2(c), pP _ Q q Ø p P Ñ Q q. By Proposition 1.1(b),
p P Ñ Q q Øp P ^ Q q. Hence, by Proposition 1.2(f),
pP _ Q q Øp P ^ Q q.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Definition
An open sentence (or proposition function) is a sentence contains
variables, which becomes a proposition when the variables are replaced by
specific mathematical objects. When P is an open sentence with a
variable x (or variables x1 , , xn ), then sentence is symbolized by P px q
(or P px1 , , xn q).
The collection of objects that can replace the variable in the open sentence
is called universe (or set of meaning) for the variable. When P px q is an
open sentence, the universe of the variable x is symbolized by X.
The set of all objects in the universe which make the open sentence a true
proposition is called the truth set.
Examples:
P(x): “x
p4, 8q.
¡ 4.”The universe can be chosen as X R. The truth set is
P(x): “x is a prime number between 5060 and 5090.”The universe can be
chosen as X N. The truth set is t5077, 5081, 5087u.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Question
Let P px q be an open sentence, and X be the universe of x. Consider the
sentences
“P px q is true for all x in X.” It is a proposition.
“P paq is true for some a in X.” It is a proposition.
For example P px q: “x ¡ 0.” and X R. Then “x ¡ 0 for all x in R.” is false,
and “a ¡ 0 for some a in R.” is true. In this case the truth set of P px q is
p0, 8q.
Remark
The truth set of an open sentence P px q depends on the universe where x
belongs to. For example, suppose that P px q is the open sentence “x 2 1 0.”
If X R, then P px q is false for all x (in X), and the truth set of P px q is empty
set. On the other hand, if X C, then P px q is true when x i, so the truth
set of P px q is ti, i u.
Notation
Let x be a variable and X be universe for x, we denote “x in X.” by x
the sentence “such that” is shorten as s.t..
Bo-Chih Huang
Foundation of Mathematics
P X.
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Definition
With a universe X specified, two open sentences P px qand Q px q are equivalent
if they have the same truth set for all x P X.
Examples:
The two sentences “3x 2 20.” and “2x 7 5.” are equivalent open
sentences in any of the number system, such as N, Z, Q, R, and C.
The two sentences “x 2 1 ¡ 0.” and “px
equivalent open sentences in R.
Bo-Chih Huang
1q ^ px ¡ 1q” are two
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Definition
The symbol D is called the existential quantifier. For an open sentence P px q,
the sentence pDx qP px q is read “there exists x s.t. P px q” or “for some x, P px q”.
Then sentence pDx qP px q is true if the truth set of P px q is non-empty.
Examples:
The quantified sentence pDx qpx 7 12x 3
universe of real numbers.
The quantified sentence (Dn)(22
universe of natural numbers.
n
16x
3 0q is true in the
1 is a prime number) is true in the
The quantified sentence (Dx, y , z, n)(x n y n z n ^ n ¥ 3) is true in the
universe of integers, but is false in the unverse of natural numbers.
Fermat Last Theorem!!
Common phrases: “there exists”, “there is”, “there are”, “for some”.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Definition
The symbol @ is called the universal quantifier. For an open sentence P px q,
the sentence p@x qP px q is read “for all x, P px q” or “for every x, P px q”, or “for
every given x (in the universe), P px q”. Then sentence p@x qP px q is true if the
truth set of P px q is the entire universe.
Examples:
The quantified sentence (@x)(x 0 _ x ¥ 0) is true in the universe of real
numbers, but is false in the universe of complex numbers.
The quantified sentence (@x, y )(x 2 y 2 ¥ 0) is true in the universe of real
numbers, but is false in the universe of complex numbers.
The quantified sentence (@f )(f is continuous on ra, b s) is true in the
universe of all differentiable real-valued functions defined on ra, b s.
Common phrases: “for each”, “for any”, “for every”, “for all”.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
In general, statement of the form “every element of the set A has the property
P” and “some element of the set A has property P” may be symbolized as
p@x P AqP px q and pDx P AqP px q, respectively. Moreover,
“All P px q are Q px q” should be symbolized as p@x qpP px q Ñ Q px qq.
“Some P px q are Q px q should be symbolized as pDx qpP px q ^ Q px qq.
Example:
(@n P N)(n is even ^ n ¡ 2 Ñ n is sum of two prime numbers)
Goldbach’s conjecture.
Definition
If P px q is a propositional function with variable x, then a counterexample to
p@x qP px q is an object t in the universe X such that P pt q is false.
Example:
P px q : p@x qpx
¡ 1q in the universe X R has counterexample x 12 .
P pnq : p@nqp22
1 is a prime numberq in the universe X N has
counterexample n 6, since
22
1 641 6700417.
n
6
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Let P px, y q be an open sentence with variables x and y and let X and Y be the
universe of x and y respectively. Consider the following sentences
“p@y
P YqpP px, y q is trueq.”
P XqpP px, y q is true).”
“p@x qp@y qpP px, y q is trueq.”
“pDy qp@x qpP px, y q is trueq.”
“pDx
It is a open sentence with variable x.
It is a open sentence with variable y .
It is a proposition.
It is a proposition.
Examples:
The sentence “For any given real number x, there is a real number y such
that x 2y .” should be symbolized as
p@x P RqpDy P Rqpx 2y q.
There exists a real number x such that x
should be symbolized as
pDx P Rqp@y P Rqpx
Bo-Chih Huang
y
2 for all real numbers y .”
y
2q .
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Remark (Hidden quantifier)
Consider the sentence
“If n is a positive integer, then n is sum of four perfect squares.”
The sentence does not appear any words like “for any”, “for all”, “there
exists” etc. We can find the hidden quantifiers by recasting the sentence:
“If n is any positive integer, then there are four perfect squares whose sum is n.”
Hence the sentence should be symbolized as
p@n P NqpDp1 , p2 , p3 , p4 P Nqpn p12
p22
p32
p42 q.
It is particularly easy to hide a universal quantifier “@”. Here is a list of some
of the more common disguises of this quantifier:
Whenever x 2 ¡ 0, whenever x ¡ 0.
If
If n is even, n2 is even.
A
A even number is multiple of 2.
For a
For a decimal to represent a fraction, it is sufficient that it be a
terminating decimal.
(All)
(All) Repeating decimals can be written as fractions.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
P X R.
p@x qp@y qpx 2 y 2 ¥ 0q. (True).
pDx qpx 2 y 2 ¡ 0q. (open sentence of y ).
p@x qp@y qpx 2 y 2 0q. (False).
pDx qpDy qpx 2 y 2 ¤ 0q. (True).
p@x qp@y qpx 2 y ¡ 0q. (False).
Counterexample: px 0q ^ py 1q.
p@x qpDy qpx y 2 ¡ 0q. (True)
Choose y 2 1 |x |.
pDy qp@x qpx y 2 ¡ 0q. (False).
Counterexample: py 1q ^ px 4q.
Examples: Let x, y
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Theorem 1.5
If P px q is a propositional function with variable x, then
a.
b.
rp@x qP px qs is equivalent to pDx q rP px qs.
rpDx qP px qs is equivalent to p@x q rP px qs.
Proof.
a. Suppose the proposition rp@x qP px qs is true. Then p@x qP px q is false, so
the truth set of P px q is not universe. Therefore, the truth set of rP px qs
is nonempty, and the proposition pDx q rP px qs is true.
Now suppose the proposition pDx q rP px qs is true. Then the truth set of
rP px qs is nonempty, so the truth set of P px q is not the universe.
Therefore, the proposition p@x qP px q is false, so rp@x qP px qs is true.
b. The proof of b., which is similar to that of part a. is left as Exercise.
Bo-Chih Huang
Foundation of Mathematics
l
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Examples:
rp@x qpDy qpDz qp@uqpDv qpx
ô pDx q rpDy qpDz qp@uqpDv qpx
ô pDx qp@y q rpDz qp@uqpDv qpx
ô pDx qp@y qp@z q rp@uqpDv qpx
ô pDx qp@y qp@z qpDuq rpDv qpx
ô pDx qp@y qp@z qpDuqp@v q rpx
ô pDx qp@y qp@z qpDuqp@v qpx y
z ¡ 2u v qs
z ¡ 2u v qs
z ¡ 2u v qs
z ¡ 2u v qs
z ¡ 2u v qs
z ¡ 2u v qs
z ¤ 2u v q
The proposition “Every positive real number has a multiplicative inverse.”
can be symbolized as p@x qpx ¡ 0 Ñ pDy qpxy 1qq. The negation of
above statement is
y
y
y
y
y
y
rp@x qpx ¡ 0 Ñ pDy qpxy 1qqs
ôpDx q rpx ¡ 0 Ñ pDy qpxy 1qqs
ôpDx qpx ¡ 0 ^ rpDy qpxy 1qsq
ôpDx qpx ¡ 0 ^ p@y q rpxy 1qsq
ôpDx qpx ¡ 0 ^ p@y qpxy 1qq
The last sentence may be translated as “There is a positive real number
that has no multiplicative inverse.”
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Examples:
Let f be defined on ra, b s, c
P pa, bq. The formal definition of xlim
Ña f px q L is
“For every ε ¡ 0 there is a δ ¡ 0 such that
if 0 |x a| δ, then |f px q L| ε.”
It can be symbolized as
p@ε ¡ 0qpDδ ¡ 0qp@x qp0 |x a| δ Ñ |f px q L| εq.
Hence, in symbols, the representation of lim f px q L is
x Ña
rp@ε ¡ 0qpDδ ¡ 0qp@x qp0 |x a| δ Ñ |f px q L| εqs
ôpDε ¡ 0q rpDδ ¡ 0qp@x qp0 |x a| δ Ñ |f px q L| εqs
ôpDε ¡ 0qp@δ ¡ 0q rp@x qp0 |x a| δ Ñ |f px q L| εqs
ôpDε ¡ 0qp@δ ¡ 0qpDx q rp0 |x a| δ Ñ |f px q L| εqs
ôpDε ¡ 0qp@δ ¡ 0qpDx qp0 |x a| δ ^ r|f px q L| εsq
ôpDε ¡ 0qp@δ ¡ 0qpDx qp0 |x a| δ ^ |f px q L| ¥ εq
That is, the meaning of lim f px q L is
x Ña
“There is a ε ¡ 0 such that for every δ ¡ 0,
there exists x P R with 0 |x a| δ but |f px q L| ¥ ε.”
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.3 Quantifiers
Definition
The symbol D! is called the unique existential quantifier. For an open
sentence P px q, the sentence pD!x qP px q is read “There is a unique x such that
P px q.” The sentence pD!x qP px q is true if the truth set of P px q has exactly one
element.
Examples:
The quantified sentence
universe N.
pD!x qpx is even and x is primeq is true in the
The quantified sentence (D!x)(x 2
in Z.
4) is true in the universe N, but false
Theorem
If P px q is a propositional function with variable x, then
a.
b.
pD!x qP px q Ñ pDx qP px q.
pD!x qP px q is equivalent to pDx qP px q ^ p@y qp@z qrP py q ^ P pz q Ñ y z s.
The proof of is left as Exercise.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Notations
N: Natural numbers (positive integers)
Z: Integers
Q: Rational numbers
R: Real numbers
C: Complex numbers
Definition
An integer n is said to be odd if Dk
P Z s.t. n 2k
An integer n is said to be even if Dk P Z s.t. n 2k.
1.
Definition
Let m, n
P Z and p P N.
n divides m, denoted by n | m if Dq P Z s.t. m nq. In this case, we say n
is a divisor(factor) of m or m is a multiple of n.
n is a perfect square if Dk
P Z s.t. n k 2 .
p is a prime if the divisors of p are only 1 and itself.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Consider a proposition “If P, then Q” or P Ñ Q. We have known that this
proposition is true under the following three situations:
Each of P and Q is true.
P is false and Q is true.
P is false and Q is false.
Since if P false, the proposition P Ñ Q is true whatever the truth value of Q
is. Therefore, in order to prove that P Ñ Q is true, it is sufficient to assume
that P is true, and under this assumption, prove that Q is true.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Direct method
Assume P
..
.
logical sequence of steps
..
.
Conclude Q.
That is, we may have to proceed the proof in several steps:
“P Ñ Q1 Ñ Q2 Ñ Ñ Qn Ñ Q”. We call this a deductive proof.
Examples:
Theorem 1.6
If n is an odd integer, then n2 is odd.
Proof:
Since n is odd, so there exists k
P Z such that n 2k
n2 p2k 1q2 4k 2 4k 1 2p2k 2
l
Since k P Z, so 2k 2 2k P Z. Hence n2 is odd.
Bo-Chih Huang
1. Then
2k q
Foundation of Mathematics
1
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Examples:
Theorem 1.7
If a, b, c
P R, then a2
b2
c2
¥ ab
bc
ca.
Proof:
Since pa b q2
so a2
b2
pb c q2 pc aq2 ¥ 0, then
2pa2 b 2 c 2 q 2pab bc
c 2 ¥ ab bc ca.
Bo-Chih Huang
caq ¥ 0
Foundation of Mathematics
l
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Contrapositive method
Assume
..
.
Q
logical sequence of steps
..
.
Conclude P.
Note that the contrapositive method is base on the tautology
pP Ñ Q q Øp Q Ñ P q.
Examples:
Theorem 1.8
Let m, b
P R with m 0 and f px q mx
Proof:
If f px q f py q, then mx
we have x y .
b
my
Bo-Chih Huang
b. If x
y , then f px q f py q.
b and hence mx
my . Since m ¥ 0, so
l
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Remark
In previous theorem, we want to prove the proposition P Ñ Q, where
P : x y and Q : f px q f py q. Without using direct method, we
assume Q : f px q f py q instead, and deduce to P : x y . By
proving Q Ñ P and use the tautology pP Ñ Q q Øp Q Ñ P q to
show the original proposition P Ñ Q is true.
Recall the proposition “P if and only if Q”, that is, P Ø Q. To prove this
proposition true, by the tautology pP Ø Q q ØrpP Ñ Q q ^ pQ Ñ P qs
(Proposition 1.2(d)), we have to prove both directions , that is, to prove
P Ñ Q is true and Q Ñ P is also true.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Examples:
Theorem 1.9
An integer n is even if and only if n2 is even.
Proof:
“Д: Recall the statement of Theorem 1.6: “If n is odd, then n2 is odd.”. The
contrapositive of this statement is “If n2 is even, then n is even.” Therefore,
the only if part is proved by Theorem 1.6.
“Ñ”: If n is even, then there exists k P Z such that n 2k. Thus
n2
Since k
P Z, so 2k 2 P Z. Hence n2 is even.
Corollary
Let n
p2k q2 4k 2 2p2k 2 q
P Z,
m
P N, then n is even if and only if nm is even.
The proof is left as an exercise.
Bo-Chih Huang
Foundation of Mathematics
l
Chapter 1. The logic and language of proofs
§1.4 Methods of proof
Examples:
Theorem 1.10
Let x
P R. Then x 1 if and only if x 3 3x 2
Proof:
“Д: If x 1, then x 3 3x 2 4x 2 1 3
“Ñ”: If x 3 3x 2 4x 2 0. Since
0 x 3 3x 2
4x
2 0.
4 2 0.
2 px 1qpx 2 2x 2q
so we get x 1 0 or x 2 2x 2 0. Since x 2 2x 2 px 1q2
so x 1 0 hence we obtain x 1.
4x
Bo-Chih Huang
Foundation of Mathematics
1 ¡ 0,
l
Chapter 1. The logic and language of proofs
§1.5 The contradiction method of proof
Contradiction method
Assume P
..
.
^
Q
logical sequence of steps
..
.
Conclude R ^ R.
Note that the contradiction method is base on the logical contradiction
R ^ R.
Examples:
Theorem 1.11
a, b, c P Z. If a, b are even and c is odd, then the equation ax
not have an integer solution for x and y .
by
c does
Proof:
Suppose ax by c has an integral solution x0 , y0 (and a, b are even, c is
odd). Then ax0 , by0 are even which implies c ax0 by0 is also even. This is a
contradiction.
l
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.5 The contradiction method of proof
Recall that a rational number r is of the form r
n 0. r is irrational if it is not rational.
Examples:
mn
for some m, n
P Z and
Theorem 1.12
?
2 is irrational.
Proof:
Suppose
?
2 P Q, then
?
2
m
for some m, n
n
P Z, n 0 and gcdpm, nq 1.
m2
Thus, 2 2 Ñ m2 2n2 , which shows that m2 is even. By Theorem 1.9, we
n
have that m is evenÑ m 2k for some k P Z. Hence
m2 2n2 Ñ n2 2k 2 Ñ n is even,
by Theorem 1.9 again. Since
? m, n are both even, then gcdpm, nq ¡ 1, a
4k 2
contradiction. Therefore,
2 is irrational.
Bo-Chih Huang
Foundation of Mathematics
l
Chapter 1. The logic and language of proofs
§1.5 The contradiction method of proof
Examples:
Theorem 1.13
There do not exist prime numbers a, b and c such that a3
b3
c 3.
Proof:
Suppose a, b, c are prime numbers s.t. a3 b 3 c 3 . If a, b are odd, then a3 , b 3
are oddÑ c 3 a3 b 3 is evenÑ c is an even primeÑ c 2. Since a, b are
odd primes, so a, b ¡ 2 but a3 b 3 8, a contradiction. Hence, a least one of
a and b is even. Assume that b 2, then
8 b3
c 3 a3 pc aqpc 2 ca a2 q
Since a, c are odd primes, so a, c ¡ 2 and thus a2 , ca, c 2 are integers ¡ 4. This
shows that c 2 ca a2 ¡ 12 which implies
8 |c 3 a3 | |c a||c 2 ca a2 | ¥ 12
This is a contradiction. Therefore, there do not exist prime numbers a, b and c
such that a3 b 3 c 3 .
l
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.5 The contradiction method of proof
Examples:
Theorem 1.14
If a, b, c are odd integers, then ax 2
solution.
Proof:
Suppose there exists p, q
0a
bx
c
0 does not have a rational
P Z, q 0 with gcdpp, qq 1 such that
p
q
2
b
p
q
c
ap
2
bpq
q2
cq 2
This shows that ap
bpq cq 0. If one of p and q is even, say p is even,
then cq 2 ap 2 bpq is also even. Since c is odd, so q 2 is even and hence q
is even, by Theorem 1.9. This implies that gcdpp, q q ¡ 1, a contradiction. A
similar case when q is even will also leads to a contradiction, which left as an
exercise. Now we can conclude that both p, q are odd. But this will deduce
that ap 2 , bpq, cq 2 are all odd and hence ap 2 bpq cq 2 is odd, a
contradiction because ap 2 bpq cq 2 0. Therefore, there is no rational
solution for ax 2 bx c 0 when a, b, c are odd.
l
2
2
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.6 More proofs
Remark
Usually a proposition can be proved using any of several different methods.
That is, the prove of a certain proposition is not unique.
Examples:
Theorem 1.15
Let x, y P R and x, y ¥ 0 such that x
12x 2 10y 2 24xy .
4y
y
3x. If 3x ¡ 2y , then
Proof (Direct method):
Note that x 4y y 3x
Ñ 4x 5y 0. If 3x 2y ¡ 0, then
p4x 5y qp3x 2y q 12x 2 23xy 10y 2 0 Ñ 12x 2 10y 2
Since x, y ¥ 0, so 12x 2 10y 2 23xy ¤ 24xy .
Bo-Chih Huang
Foundation of Mathematics
23xy .
l
Chapter 1. The logic and language of proofs
§1.6 More proofs
Remark
Usually a proposition can be proved using any of several different methods.
That is, the prove of a certain proposition is not unique.
Examples:
Theorem 1.15
Let x, y P R and x, y ¥ 0 such that x
12x 2 10y 2 24xy .
4y
y
3x. If 3x ¡ 2y , then
Proof (Contrapositive method):
Suppose 12x 2 10y 2 ¥ 24xy . Since x, y ¥ 0, then
12x 2 10y 2 ¥ 24xy ¥ 23xy . Also, since x 4y y 3x, so 4x
Thus
p4x 5y qp3x 2y q 12x 2 10y 2 23xy ¥ 0.
and hence 3x
2y ¤ 0.
Bo-Chih Huang
Foundation of Mathematics
5y
0.
l
Chapter 1. The logic and language of proofs
§1.6 More proofs
Remark
Usually a proposition can be proved using any of several different methods.
That is, the prove of a certain proposition is not unique.
Examples:
Theorem 1.15
Let x, y P R and x, y ¥ 0 such that x
12x 2 10y 2 24xy .
4y
y
3x. If 3x ¡ 2y , then
Proof (Contradiction method):
Suppose that 3x ¡ 2y and 12x 2 10y 2 ¥ 24xy . Since
x 4y y 3x Ñ 4x 5y 0 and 3x 2y ¡ 0, then
0 ¡ p4x
5y qp3x 2y q 12x 2
10y 2 23xy
¥ 24xy 23xy xy ¥ 0,
l
a contradiction.
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.6 More proofs
Examples:
Theorem 1.16 (Proof by cases)
If x is any real number such that x 2 x
2
0, then
1
x
2.
Proof:
Choose x P R s.t. x 2 x 2 px 2qpx 1q 0. Then either the following
cases will happen: px 2 ¡ 0q ^ px 1 0q or px 2 0q ^ px 1 ¡ 0q.
Case 1. If x 2 ¡ 0 and x 1 0, then x ¡ 2 and x 1. There is no such
real number.
Case 2. If x 2 0 and x 1 ¡ 0, then 1 x 2.
Since the proof is independent of the real number x chosen, as long as it
satisfies the condition x 2 x 2 0. Hence we have prove the Theorem. l
Bo-Chih Huang
Foundation of Mathematics
Chapter 1. The logic and language of proofs
§1.6 More proofs
Examples:
Theorem 1.17
There do not exist natural numbers m and n such that
7
17
m1
1
.
n
Proof:
Let m, n P N. Denote mintm, nu be the smaller number in m and n, and
maxtm.nu be the larger number in m and n.
Claim: mintm, nu ¡ 2: If not, then either m ¤ 2 or n ¤ 2. Thus
7
m1 n1 ¥ 12 , a contradiction.
17
7
m1 n1 ¤
Claim: mintm, nu 5: If not, then both m, n ¥ 5. Thus
17
contradiction.
1
Hence one of m and n must be 3 or 4. But 7
514 R N and
1
17
3
1
68
R
N,
which
both
leads
to
contradiction.
Therefore, there do
7
14 11
17
7
exist m, n P N such that
m1 n1 .
17
Bo-Chih Huang
Foundation of Mathematics
2
,a
5
not
l
Chapter 1. The logic and language of proofs
§1.6 More proofs
Question
No!
Is mathematics just a game of logic?
Examples:
Theorem 1.18
For x, y
P R and x
y
π, then sinpx
y q sin x cos y
cos x sin y .
Proof:
a cos y b cos x (Pythagorean theorem)
c
1
r Ñ sin z 21 c sin x a, sin y b
2
c
2
(Law of Sine)
sinpx
y q sinpπ px
y qq
sin z sin x cos y
cos x sin y
l
Bo-Chih Huang
Foundation of Mathematics