Proof methods- teachers

Section 4: Proof methods
SQU-Math2350
Dr. Yassir Dinar
Sultan Qaboos University
Department of Mathematics
Math2350: Foundation of Mathematics
Fall 2019
SQU-Math2350 Section 4: Proof methods Fall 2019 1 / 15

Structure of Mathematics: Informal definitions

Definition 4.1
Undefined terms are known concepts fundamentals to the context of
study which are not defined formally.

Definition 4.1
Undefined terms are known concepts fundamentals to the context of
study which are not defined formally.
Example 4.2
In this course undefined concepts include natural, integers and real numbers.
They are known but are undefined formally.

Axioms

Axioms
Definition 4.3
Axioms or postulates are statements and propositions that are
assumed to be true.
Example 4.4
In this course must properties of natural, integer and real numbers are assumed to
be true like

Axioms
Definition 4.3
assumed to be true.
Example 4.4
be true like
1 The product and sum of integers is an integer.

Axioms
Definition 4.3
assumed to be true.
Example 4.4
be true like
2 Let a,b be real numbers. If ab = 0 then a = 0 or b = 0.

Axioms
Definition 4.3
assumed to be true.
Example 4.4
be true like
3 Let a,b and c be real numbers and c > 0. Then ac > bc if and only if a > b.

Axioms
Definition 4.3
assumed to be true.
Example 4.4
be true like
3 Let a,b and c be real numbers and c > 0. Then ac > bc if and only if a > b.
4 All properties in the appedix II of the reference book are assumed to be true.

Definitions

Definitions
Definition 4.5
A definition is used to introduce new concepts using axioms and undefined
terms.

Definitions
Definition 4.5
terms.
Example 4.6
Here are examples of new concepts introduced earlier

Definitions
Definition 4.5
terms.
Example 4.6
1 Divisibility on the set of integers.

Definitions
Definition 4.5
terms.
Example 4.6
2 Prime and composite numbers in N.

Definitions
Definition 4.5
terms.
Example 4.6
3 Even and odd integers.

Definitions
Definition 4.5
terms.
Example 4.6
4 Rational and irrational numbers in R.

Definitions
Definition 4.5
terms.
Example 4.6
4 Rational and irrational numbers in R.
5 Continuous functions on the set of real functions.

Theorems and Proofs

Theorems and Proofs
Definition 4.7
A theorem is a proposition that describes a pattern or relationship
among quantities or structures which can be proved to be true by using
principles of logic.

Theorems and Proofs
Definition 4.7
principles of logic.A proof of a theorem is a justification of the truth of
the theorem that follow the principle of logic.
Example 4.8
This course gives tools to prove statements like:

Theorems and Proofs
Definition 4.7
Example 4.8
1 Let a and b be two integers: if a and b are even then a + b is even.

Theorems and Proofs
Definition 4.7
Example 4.8
2 There exist a real number x such that x4
− 4x + 2 = 0.

Theorems and Proofs
Definition 4.7
Example 4.8
− 4x + 2 = 0.
3 Let a and b be two integers: ab is even if and only if a is even or b is even.

Theorems and Proofs
Definition 4.7
Example 4.8
− 4x + 2 = 0.
4 Product of rational and irrational numbers are irrational.

Theorems and Proofs
Definition 4.7
Example 4.8
− 4x + 2 = 0.
5 Let x and y be positive real numbers such that x − 4y < y − 3x. Prove that
if 3x > 2y then 12x2
+ 10y2
< 24xy.

Theorems and Proofs
Definition 4.7
Example 4.8
− 4x + 2 = 0.
5 Let x and y be positive real numbers such that x − 4y < y − 3x. Prove that
if 3x > 2y then 12x2
+ 10y2
< 24xy.
6
√
2 is irrational number.

Modus Ponens
One of the basic tools in proofs

Modus Ponens
Definition 4.9
Modus ponens rule states that when both proposition P and P ⇒ Q are
true then Q must be true.

Modus Ponens
Definition 4.9
Proof.
From the truth table of conditional statement when P is true, then
P ⇒ Q is true only when Q is true.

Modus Ponens
Definition 4.9
Proof.
Example 4.10
Let n, m be integers. Show that 3n − 7m is an integer.

Modus Ponens
Definition 4.9
Proof.
Example 4.10
Solution.
It is known that sum and product of integers is an integer.

Modus Ponens
Definition 4.9
Proof.
Example 4.10
Solution.
It is known that sum and product of integers is an integer. Since m and n
are integers.

Modus Ponens
Definition 4.9
Proof.
Example 4.10
Solution.
It is known that sum and product of integers is an integer. Since m and n
are integers. then 3n − 7m is an integer (by modus ponens).

Direct Proof

Direct Proof
One of the basic proofs methods for statement having the proposition
form P ⇒ Q.

Direct Proof
form P ⇒ Q.
Direct Proof of P ⇒ Q.
Assume P.

Direct Proof
form P ⇒ Q.
Assume P.
{Use definitions, axioms, properties,algebraic manipulation,
principles of logic, mathematical tricks, etc...}

Direct Proof
form P ⇒ Q.
Assume P.
{Use definitions, axioms, properties,algebraic manipulation,
principles of logic, mathematical tricks, etc...}
Therefore Q,
Thus, P ⇒ Q.

Examples of Direct Proof I

Example 4.11
Let x be an integer. Prove that if x is odd, then x + 1 is even.

Example 4.11
Solution.

Example 4.11
Solution.
Let x be an integer. Suppose x is odd.

Example 4.11
Solution.
Let x be an integer. Suppose x is odd.Then, by definition, x = 2k + 1 for
some integer k.

Example 4.11
Solution.
some integer k.
x + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1).

Example 4.11
Solution.
some integer k.
x + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1).
Since k is an integer. Then k + 1 is an integer.

Example 4.11
Solution.
some integer k.
x + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1).
Since k is an integer. Then k + 1 is an integer. Then x + 1 is a product of
2 and the integer k + 1. Then, by definition, x + 1 is even.

Example 4.11
Solution.
some integer k.
x + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1).
Since k is an integer. Then k + 1 is an integer. Then x + 1 is a product of
2 and the integer k + 1. Then, by definition, x + 1 is even.Therefore, if x
is odd, then x + 1 is even.

Examples of Direct Proof II

Example 4.12
Let a, b and c be integers. If a divides b and b divides c, then a divides c.

Example 4.12
Solution.
Let a, b and c be integers. Suppose a divides b and b divides c.

Example 4.12
Solution.
Let a, b and c be integers. Suppose a divides b and b divides c. Then, by
definition, b = ak for some integer k and c = bm for some integer m.

Example 4.12
Solution.
definition, b = ak for some integer k and c = bm for some integer m. Then
c = bm = (ak)m = a(km).

Example 4.12
Solution.
c = bm = (ak)m = a(km).
Since k and m are integers, km is an integer. Then, by definition, a
divides c.

Example 4.12
Solution.
c = bm = (ak)m = a(km).
Since k and m are integers, km is an integer. Then, by definition, a
divides c. Therefore, If a divides b and b divides c, then a divides c.

Examples of Direct Proof III, (Working Backward)

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Note that form the context , x is a real number.

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0
Now (x − 5)(x − 2) > 0 if both (x − 5) and (x − 2) are positive or both
are negative.

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0
are negative.
x2
≤ 1 ⇒ −1 ≤ x ≤ 1 ⇒ x ≤ 1 ⇒ x < 2 ∧ x < 5
⇒ [(x − 2) < 0] ∧ [(x − 5) < 0].

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0
are negative.
x2
≤ 1 ⇒ −1 ≤ x ≤ 1 ⇒ x ≤ 1 ⇒ x < 2 ∧ x < 5
⇒ [(x − 2) < 0] ∧ [(x − 5) < 0].
Solution.
Assume x2 ≤ 1 .

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0
are negative.
x2
≤ 1 ⇒ −1 ≤ x ≤ 1 ⇒ x ≤ 1 ⇒ x < 2 ∧ x < 5
⇒ [(x − 2) < 0] ∧ [(x − 5) < 0].
Solution.
Assume x2 ≤ 1 . Then −1 ≤ x ≤ 1. Then x ≤ 1.

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0
are negative.
x2
≤ 1 ⇒ −1 ≤ x ≤ 1 ⇒ x ≤ 1 ⇒ x < 2 ∧ x < 5
⇒ [(x − 2) < 0] ∧ [(x − 5) < 0].
Solution.
Assume x2 ≤ 1 . Then −1 ≤ x ≤ 1. Then x ≤ 1.Thus x < 2 and x < 5,
and so (x − 2) < 0 and (x − 5) < 0.

Example 4.13
Prove that if x2
≤ 1, then x2
− 7x > −10.
Working Backward:
x2
− 7x > 10 ⇔ x2
− 7x − 10 > 0 ⇔ (x − 5)(x − 2) > 0
are negative.
x2
≤ 1 ⇒ −1 ≤ x ≤ 1 ⇒ x ≤ 1 ⇒ x < 2 ∧ x < 5
⇒ [(x − 2) < 0] ∧ [(x − 5) < 0].
Solution.
Assume x2 ≤ 1 . Then −1 ≤ x ≤ 1. Then x ≤ 1.Thus x < 2 and x < 5,
and so (x − 2) < 0 and (x − 5) < 0. Then,
(x − 5)(x − 2) = x2 − 7x − 10 > 0. Therefore, x2 − 7x > 10 .

Examples of Direct Proof IV (by Exhaustion)
A proof by exhaustion consists of an examination of every possible case.

Example 4.14
If the statement contains |x|, then two cases can be considered, x ≥ 0 which
implies |x| = x or x < 0 which implies |x| = −x.

Example 4.14
Example 4.15
Let x be a real number. Prove that −|x| ≤ x ≤ |x|.

Example 4.14
Example 4.15
Solution.
Let x be a real number:

Example 4.14
Example 4.15
Solution.
Case 1: Suppose x ≥ 0. Then |x| = x.

Example 4.14
Example 4.15
Solution.
Case 1: Suppose x ≥ 0. Then |x| = x. Since x ≥ 0, −x ≤ x.

Example 4.14
Example 4.15
Solution.
Case 1: Suppose x ≥ 0. Then |x| = x. Since x ≥ 0, −x ≤ x. Hence
−x ≤ x ≤ x which is −|x| ≤ x ≤ |x|.

Example 4.14
Example 4.15
Solution.
−x ≤ x ≤ x which is −|x| ≤ x ≤ |x|.
Case 2: Suppose x < 0. Then |x| = −x.

Example 4.14
Example 4.15
Solution.
−x ≤ x ≤ x which is −|x| ≤ x ≤ |x|.
Case 2: Suppose x < 0. Then |x| = −x. Since x < 0, x ≤ −x.

Example 4.14
Example 4.15
Solution.
−x ≤ x ≤ x which is −|x| ≤ x ≤ |x|.
Hence x ≤ x ≤ −x which is −|x| ≤ x ≤ |x|.

Example 4.14
Example 4.15
Solution.
−x ≤ x ≤ x which is −|x| ≤ x ≤ |x|.
Hence x ≤ x ≤ −x which is −|x| ≤ x ≤ |x|.
Thus in both cases −|x| ≤ x ≤ |x|.

Without loss of generality

The expression “ Without loss of generality ” is used when a proof by
exhaustion leads to cases so similar in reasoning.

Example 4.16
Let m and n be integers. Prove that if one of them is odd and the other is even,
then m2
+ n2
= 4k + 1 for some integer k.

Example 4.16
then m2
+ n2
Solution.
Let m and n be integers. Without loss of generality, assume m is even and
n is odd.

Example 4.16
then m2
+ n2
Solution.
n is odd. Then, by definition,

Example 4.16
then m2
+ n2
Solution.
n is odd. Then, by definition, there exist integers t,u such that m = 2t
and n = 2u + 1. Then
m2
+ n2
= 4(t2
+ u2
+ u) + 1
Since t2 + u2 + u is an integer, m2 + n2 has the form 4k + 1 for some
integer k

Direct Proofs of Compound Propositions

Form Proof Outline reason

(P ∧ Q) ⇒ R Assume P and Q · · · . Thus R -

P ⇒ (Q ∧ R) Assume P · · · . Thus Q · · · thus R -

(P ⇒ Q) ⇒ R Assume P ⇒ Q · · · . Thus R -

(P ∨ Q) ⇒ R Case 1: Assume P · · · . Thus R P ∨ Q ⇒ R ≡
Case 2: Assume Q · · · . Thus R (P ⇒ R) ∧ (Q ⇒ R)

P ⇒ (Q ∨ R) Assume P and ∼ Q · · · . Thus R P ⇒ (Q ∨ R) ≡
(P∧ ∼ Q) ⇒ R

P ⇒ (Q ∨ R) Assume P and ∼ Q · · · . Thus R P ⇒ (Q ∨ R) ≡
(P∧ ∼ Q) ⇒ R
P ⇒ (Q ⇒ R) Assume P and Q · · · . Thus R P ⇒ (Q) ≡
(P ∧ Q) ⇒ R

Proofs Outlines I

Proofs Outlines I
Example 4.17
Let f be a polynomial. Outline a direct proof that if f has degree 4, then f
has a real zero or f is product of two irreducible quadratics.

Proofs Outlines I
Example 4.17
Solution.
The statement is in the form P ⇒ (Q ∨ R).

Proofs Outlines I
Example 4.17
Solution.
The statement is in the form P ⇒ (Q ∨ R). A possible outline is:
Assume f has degree 4 and f has no real zero
.
.
.

Proofs Outlines I
Example 4.17
Solution.
The statement is in the form P ⇒ (Q ∨ R). A possible outline is:
Assume f has degree 4 and f has no real zero
.
.
.
Therefore, f is a product of two irreducible quadratics.

Proofs Outlines II

Proofs Outlines II
Example 4.18
Outline a direct proof that if a quadrilateral has opposite sides equal or opposite
angles equal, then it is a parrellelogram.

Proofs Outlines II
Example 4.18
Solution.
Let a be a quadrilateral.
Case 1: Assume a has equal opposite sides
.
.
.

Proofs Outlines II
Example 4.18
Solution.
.
.
.
Therefore, a is a parallelogram.

Proofs Outlines II
Example 4.18
Solution.
.
.
.
case 2: Assume a has equal opposite angles
.
.
.

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