Methods of Proof_Handouts

METHODS OF PROOF

METHODS OF PROOF

RICKY F. RULETE

Department of Mathematics and StatisticsUniversity of Southeastern Philippines

METHODS OF PROOF

VALID ARGUMENTS

Basic Valid Argument Forms

Direct Implication

p qp

q

Contrapositive Implication

p qq p

METHODS OF PROOF

VALID ARGUMENTS


Transitivity of

p qq r p r

Two Separate Cases

p rq rp q r

METHODS OF PROOF

VALID ARGUMENTS


Eliminating a Possibility

p qp q

In Particular

p q p

METHODS OF PROOF

VALID ARGUMENTS


Obtaining Or

p

p q

Obtaining And

p

q

p q

METHODS OF PROOF

VALID ARGUMENTS


Substitution of Equivalent

p qp

q

METHODS OF PROOF

ARGUMENTS INVOLVING QUANTIFIERS

Principle of Specification

If the premises

x U , p(x) and a U

hold, then the conclusionp(a)

also holds.

METHODS OF PROOF

ARGUMENTS INVOLVING QUANTIFIERS

Principle of Generalization

From the following steps:

1 Take an arbitrary element a U .2 Establish that p(a) holds.

the conclusion x U , p(x)

is obtained.

METHODS OF PROOF

Direct Proof

Example (1)

Show that the points (4,5), (2,2), and (8, 1) lie on a commonline.

Proof.

Let L be the line given by the equation y = 12x 3. Observe that

5 = 12(4) 3,

2 = 12(2) 3, and

1 =12(8) 3.

Therefore, all of the points (4,5), (2,2), and (8, 1) lie on acommon line L.

METHODS OF PROOF

Direct Proof

The symbol signifies the end of the proof. Some people insteaduse Q.E.D., which stands for the Latin phrase quod eratdemonstrandum and means which was to be demonstrated.

METHODS OF PROOF

Direct Proof

Existential Statements

To prove a statement of the form

x U such that p(x)

it suffices to present an example of a particular element x U forwhich p(x) holds.

METHODS OF PROOF

Direct Proof

Example (2)

Show: There is a set A such that A R+ = Z+.

Proof.

Let A = Z. Observe that

A R+ = Z R+ = Z+.

METHODS OF PROOF

Direct Proof

Example (3)

Show: There exist sets A and B such that |A B| < |A|+ |B|.

Proof.

Let A = {1, 2} and B = {2, 3}. So A B = {1, 2, 3}. Observe that

|A B| = 3 < 2 + 2 = |A|+ |B|.

METHODS OF PROOF

Counterexamples

Example (3)

Disprove: Every intervals complement is not an interval.

Counterexample

Let I = [0,). So I is an interval. Observe that

Ic = [0,)c = (, 0).

So Ic is also an interval. Hence, there is an interval whosecomplement is also an interval.

METHODS OF PROOF

Counterexamples

Example (4)

Prove or Disprove: x R, if x < 2, then x2 < 4.

Counterexample

Let x = 3. Then x2 = 9. Observe that x < 2 and x2 4. That is,for x = 3, it is not true that

if x < 2, then x2 < 4.

METHODS OF PROOF

Counterexamples

Example (5)

Show: x R, x2 + 1 > 0.

Proof.

Let x R. Since the square of any real number is nonnegative, wehave x2 0. Hence,

x2 + 1 0 + 1 = 1 > 0.

We therefore have x2 + 1 > 0.

METHODS OF PROOF

Counterexamples

Example (6)

Show: x R, if x [2, 3], then x2 [4, 9].

Proof.

Let x R and suppose that x [2, 3]. That is, 2 x 3. Squaringeach term gives 22 x2 32. Hence, 4 x2 9. Therefore,x [4, 9].

METHODS OF PROOF

Counterexamples

Example (7)

Show: If x is an odd integer, then x2 is odd.

Proof.

If x is odd, then x = 2a+ 1 for some integer a. Now let us look at x2.We have

x2 = (2a+ 1)2 = 4a2 + 4a+ 1 = 2(2a2 + 2a) + 1.

Since a is an integer, it follows that 2a2 + 2a+ 1 is also an integer.Hence, x2 is odd.

METHODS OF PROOF

Indirect Proof

Proof by Contradiction

Example (8)

Show: If x is an odd integer, then x2 is odd.

METHODS OF PROOF

Indirect Proof


Proof.

Suppose that x is odd and x2 is even. Then x = 2a+ 1, and x2 = 2bfor some integers a and b. Hence

2b = x2 = (2a+ 1)2 = 4a2 + 4a+ 1 = 2(2a2 + 2a) + 1.

Hence,1 = 2b 2(2a2 + 2a) = 2[b (2a2 + 2a)].

But [b (2a2 + 2a)] is clearly an integer, so the last equation impliesthat 1 is divisible by 2 with the integer [b (2a2 + 2a)] as quotient.Hence, we cannot assume that the assertion is false. Therefore, it mustbe true.

METHODS OF PROOF

Indirect Proof


Example (9)

Show: R+ does not have a smallest element.

Proof.

Suppose R+ has a smallest element, say s. Then s x for everyx R+. However, s2 is a smaller element of R+ since s2 < s ands2 > 0. This contradicts the fact that s was supposed to be the smallestelement.

METHODS OF PROOF

Indirect Proof


Example (10)

Show: Z is infinite.

Proof.

Suppose Z is finite. Let n be the number of elements of Z. However,there are n+ 1 distinct integers in the list 1, 2, 3, . . ., n, n+ 1. SoZhas more than n elements. This is a contradiction.

METHODS OF PROOF

Indirect Proof

Proof by Contrapositive

Example (11)

Show: If the sum of two real numbers is positive, then at least one ofthem is positive, i.e., if x, y R, and x+ y > 0, then either x > 0 ory > 0.

Proof.

Assume that x 0 and y 0. Then x+ y x+ 0 = x 0. Thusx+ y 0, which means that the hypothesis is false. This proves thecontrapositive, hence the assertion is true.

METHODS OF PROOF

Splitting into Cases

Example (12)

Show: x R, if |x| > 1, then x2 > 1.

Proof.

Suppose x R and that |x| > 1. So x > 1 (when x 0) or x > 1(when x < 0).

Case 1: If x > 1, then x2 > 12 = 1.

Case 2: If x > 1, then (x2 > 12 = 1). Since(x)2 = (1)2x2 = x2, substitution gives that x2 > 1.

In both cases, x2 > 1.

METHODS OF PROOF


Example (13)

Let a, b, c R with a 6= 0. Prove that the number of distinct real rootsof the quadratic polynomial

ax2 + bx+ c is

2 if b2 4ac > 01 if b2 4ac = 00 if b2 4ac < 0.

METHODS OF PROOF


Proof

The roots of the equation

ax2 + bx+ c = 0

are given by the quadratic formula

x =bb2 4ac

2a.

Case 1: If b2 4ac > 0, then the roots are

b+b2 4ac2a

andbb2 4ac

2a.

Moreover, these are distinct.

METHODS OF PROOF


Proof

Case 2: If b2 4ac = 0, then the single root is b2a .Case 3: If b2 4ac < 0, thenb2 4ac does not exist as a real

number. So there are no real roots.

METHODS OF PROOF

MATHEMATICAL INDUCTION

Principle of Mathematical Induction I:

Let P be a proposition defined on the set of positive integers Z+; thatis, P(n) is either true or false for each n Z+. Suppose P has thefollowing two properties:

1 P(1) is true.

2 P(k + 1) is true whenever P(k) is true.

The P is true for every positive integer n Z+.

METHODS OF PROOF


Example (14)

Let P be the proposition that the sum of the first n positive oddnumbers is n2; that is,

P(n) : 1 + 3 + 5 + + (2n 1) = n2

Observe that P(n) is true for n = 1;namely,

P(1) : 1 = 12

METHODS OF PROOF


Example (14)

Assuming P(k) is true, we add 2k + 1 to both sides of P(k), obtaining

1 + 3 + 5 + + (2k 1) + (2k + 1) = k2 + (2k + 1)= (k + 1)2

which is P(k + 1). In other words, P(k + 1) is true whenever P(k) istrue. By the principle of mathematical induction, P is true for all n.

METHODS OF PROOF


Remark

Sometimes one wants to prove that a proposition P is true for the setof integers

{a, a+ 1, a+ 2, a+ 3, . . .}where a is any integer, possibly zero. This can be done by simply

replacing 1 by a in either of the above Principles of MathematicalInduction.

METHODS OF PROOF


Example (15)

Show: n 4, n2 3n+ 4.

Proof

Note that 42 = 16 3(4) + 4. Suppose k 4 and that k2 3k + 4.Observe that

(k + 1)2 = k2 + (2k + 1)

(3k + 4) + (2k + 1) By the inductive hypothesis= 3k + (2k + 5)

3k + 7.

That is, (k + 1)2 3(k + 1) + 4.

METHODS OF PROOF


Principle of Mathematical Induction II:

Let P be a proposition defined on the set of positive integers Z+ suchthat:

1 P(1) is true.

2 P(k) is true whenever P(j) is true for all 1 j < k.The P is true for every positive integer n Z+.

METHODS OF PROOF


The sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . is known asthe Fibonacci sequence. It is named after the Italian mathematicianLeonardo of Pisa (known as Fibonacci) who lived in the late twelfthand early thirteenth centuries. If we denote the Fibonacci sequenceby {Fn}n0, then

F0 = 1,F1 = 1, and n 2,Fn = Fn2 + Fn1.

METHODS OF PROOF


Example (16)

Show that the Fibonacci sequence can be expressed by the formula

n 0,Fn = 15

(1 +52

)n+1(

152

)n+1 .

METHODS OF PROOF


Proof

It is straightforward to check that

15

(1 +52

)1(

152

)1 = 1, and15

(1 +52

)2(

152

)2 = 1.

METHODS OF PROOF


Suppose k 1 and that

Fi =15

(1 +52

)i+1(

152

)i+1 ,for each 0 i k. Note that k + 1 2 and that both k 1 and k lie

in the interval [0, k].

METHODS OF PROOF


Observe that

Fk+1 = Fk1 + Fk

=15

(1 +52

)k(

152

)k+

15

(1 +52

)k+1(

152

)k+1=

15

(1 +52

)k+

(1 +

52

)k+1

(15

2

)k(

152

)k+1

METHODS OF PROOF


Fk+1 =15

(1 +52

)k(1 +

1 +

52

)

(15

2

)k(1 +

152

)

METHODS OF PROOF


Fk+1 =15

(1 +52

)k(3 +

52

)

(15

2

)k(35

2

)

METHODS OF PROOF


Fk+1 =15

(1 +52

)k(1 +

52

)2

(15

2

)k(15

2

)2

METHODS OF PROOF


Fk+1 =15

(1 +52

)k+2(

152

)k+2

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