Lecture 3Method of proofs

Consider the statements: “Humans have two eyes”

It implies the “universal quantification” If a is a Human then a has two eyes.

Introduction

Mathematical theorem:

“If x > y, where x and y are positive real numbers, then x2 > y2”

It stated that, if “x > y” is TRUE, then “x2 > y2” must be TRUE and the case where “x2 > y2” is FALSE will NEVER occurs.

How to prove this statement is valid?

Introduction

A definition is an exact, unambiguous explanation of the meaning of a mathematical word or phrase. (We do not need to prove definition)◦ E.g. An integer n is even if n = 2a for some

integer a 2 Z. A theorem is a statement that is true and

has been proved to be true.◦ E.g.: Every absolutely convergent series

converges.

Concepts

Definition 3.The integer n is even if there exists an integer k such that n = 2k and it is odd if there exists an integer k such that n = 2k + 1.Example:6 is an even integer, because you can find k=3 such that 6 = 2 3.9 is an odd integer, because you can find k=4 such that 9 = 2 4 + 1.

Definitions of odd and even Integers

A proof of a theorem is a written verification

that shows that the theorem is definitely and certainly true.

What is a proof?

Direct proof Indirect proof Proof by contradiction

Proof methods

P Q

Shows that if P is true, then Q must be true.Suppose that P is true...Therefore, Q is true.

Direct Proof

Give a direct proof of the theorem “If n is odd, then n2 is odd”

 Solution:Suppose that n is odd.Then n = 2k + 1, where k (using definition 3).If follows that n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1=2b + 1, where b = 2k2 + 2k Therefore, n2 is odd.

Direct Proof

If x is an even integer, then x2- 6x + 5 is odd.

Proof:

P Q is logically equivalent to its contrapositive

P Q QP

So, in indirect proof, instead of proving P Q, we prove QP.

E.g.Prove that “If n is odd, then n2 is odd”

Prove that “If n2 is even, then n is even”

Indirect Proof

P Q

Steps:

Suppose that Q (is true)

.

.

.

Therefore P.

Indirect Proof

Example:Given an indirect proof of the theorem “If 3n + 2 is odd, then n is odd.”

Solution:The contrapositive is, “If n is even, then 3n + 2 is even”.Suppose n is even. Then n = 2k where k .It follows that 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k +1)=2b, where b = 3k+1 . So 3n + 2 is even (since it is multiple of 2).

Indirect Proof

Prove that if n2 is even, then n is even. The contrapositive of this statement is “if n is odd, then n2 is odd”

 Let n be an odd integer. Since n is odd, there exists an integer k so that n=2k + 1

Squaring both sides, we haven2 = (2k + 1)2

= 4k2 + 4k + 1=2(2k2 + 2k) + 1

Suppose J= 2k2 + 2k , we haven2 = 2J + 1By definition of odd integers, n2 is odd, and the proposition is proved.

Another example

Proof by contradiction

If a man accused of holding up a bank can prove that he was some place else at the time the crime was committed, he will certainly be acquitted.

Assume that he committed the crime. Then at the time of the crime, I would have

had to be at the scene of the crime.

Crime case

In fact, at the time of the crime I was in a meeting with 20 people far from the crime scene, as they will testify.

This contradicts the assumption that I committed the crime, since it is impossible to be in two places at one time. hence that assumption is false.

(continue)

To prove that a statement p is true. Suppose that p is false (p is true). Show that it produces contradictory results

c c. So the assumption of p must be false.

Proof by contradiction

Prove the theorem “There is no greatest integer”.

Proof: Suppose the opposite is true there is a greatest

integer N. Then N ³ n for every integer n. Let M = N + 1. Now M is an integer since it is a sum of integers. Also M > N since M = N + 1.

Thus M is an integer that is greater than N. So N is the greatest integer and N is not the greatest integer, which is a contradiction.

Example

Given the implication pq, suppose that the conclusion q is false (q and p is true),

following a sequence of steps, we deduce the statement p (false).

But p is a contradiction to the supposition that p and q are true. Therefore q must be true.

Proof by contradiction for pq

Prove that “If 3n + 2 is odd, then n is odd”

Example

Arguing from examples◦ E.g: The sum of any two even integers is even.

Solution: This is true because if m = 14 and n = 6, which are

both even, then m + n = 20, which is also even.

It is not sufficient to show that the conclusion “m + n is even” is true for m=14 and n = 6. You must give an argument to show that the conclusion is true for any even integers m and n.

Mistakes in proof

Using the same letter to mean two different things.

Consider the following proof:

Suppose m and n are odd integers. Then by definition of odd, m = 2k + 1 and n = 2k + 1 for some integer k.

This is incorrect. Using the same symbol, k, in the expressions for both m and n implies that m = 2k + 1 = n.

Mistakes in proof

Jumping to a conclusion.◦ Jump to conclusion without giving an adequate

reason. Consider the following proof that the sum of any two even integers is even.

Suppose m and n are any even integers. By definition of even, m=2r and n = 2s for some integers r and s. Then m + n = 2r and 2s. So m + n is even.

The problem with this proof is the crucial calculation 2r + 2s = 2 (r + s) is missing.

Mistakes in proof

Begging the question◦ Assume what is to be proved.

Consider: Prove that the product of any two odd

integers is odd.

Mistakes in proof

Suppose that m and n are odd integers. If mn is odd, then mn=2k + 1 for some integer k. Also by definition of odd, m = 2a + 1 and n = 2b + 1 for some integers a and b. Then mn = (2a + 1)(2b + 1) = 2k + 1, which is odd by definition of odd.

It is wrong because it assumes that the conclusion mn is odd is true, and later assumes it to be true by setting (2a + 1)(2b + 1)=2k + 1.

Mistakes in proof (continue)

To disprove xP(x) Q(x) is false is to equivalent to show that

xP(x) Q(x)

For example, show that “if n is even then n2 is even”

If we can find at least an “even” number which square is “not even” then this statement is false.

The example that show a universal statement is false is called the counterexample

Disproving Universal Statements by Counterexample

Disprove the following statement by finding a counterexample:

a,b R, if a2 = b2 then a = b.

Solution:To disprove this statement, you need to find real numbers a and b such that a2=b2 and a b. For example, when a =-1 and b=1.

Example

We wish to establish the truth of xP(x). That is, P(x) is true for at least one x in the domain.

One way to prove this statement is to find an x that makes P(x) is true.

Then, generalize the results using Existential Generalization rule.

Proving Existential Statement

Existential generalization (EG)

If Q(x) is true, then there xQ(x) is true.

Proving Existential Statement

Theorem: There exists an integer solution to the equation x2 + y2 = z2.

Proof:Choose x = 3, y = 4, z = 5 so that 32 + 42=52. According to EG, there exists integers such that x2 + y2 = z2.

Example

Proof techniques are useful to prove or disprove the validity of a statement

Summary