lecture 3 – february 17, 2003. chapter 3 elementary number theory and methods of proof

Lecture 3 – February 17, 2003

Chapter 3

Elementary Number Theory and Methods of Proof

Section 3.1

Direct Proof and Counterexample I: Introduction

Definitions A definition gives meaning to a term. A non-primitive term is defined using previously

defined terms. A primitive term is undefined. Example

A function f : R R is increasing if f(x) f(y) whenever x y.

Previously defined terms: function, real numbers, greater than.

Definitions Definitions are not theorems. Example

Def: A number n is a perfect square if n = k2 for some integer k.

Now suppose t is a perfect square. Then t = k2 for some integer k. Is this the “error of the converse”?

Definitions are automatically “if and only if,” even though they don’t say so.

Proofs A proof is an argument leading from a

hypothesis to a conclusion in which each step is so simple that its validity is beyond doubt.

That is a subjective judgment – what is simple to one person may not be so simple to another.

Types of Proofs Proving universal statements

Prove something is true in every instance Proving existential statements

Prove something is true in at least one instance Disproving universal statements

Prove something is false in at least one instance Disproving existential statements

Prove something is false in every instance

Proving Universal Statements The statement is generally of the form

x D, P(x) Q(x) Use the method of generalizing from the

generic particular. Select an arbitrary x in D (generic particular). Assume that P(x) is true (hypothesis). Argue that Q(x) is true (conclusion).

Theorem: The sum of two consecutive triangle numbers is a perfect square. Definition: Let n be a positive integer. The nth

triangle number Tn is the number n(n + 1)/2.

Definition: Let n be a positive integer. The nth perfect square Sn is the number n2.

Example: Direct Proof

Proof of Theorem Proof:

Let n be a positive integer. Tn + Tn + 1 = n(n + 1)/2 + (n + 1)(n + 2)/2

= (n2 + n + n2 + 3n + 2)/2= (2n2 + 4n + 2)/2= (n + 1)2

= Sn + 1. Therefore, Tn + Tn + 1 = Sn + 1 for all n 1.

Example: Direct Proof Theorem: If x, y R, then

x2 + y2 2xy. Incorrect proof:

x2 + y2 2xy. x2 – 2xy + y2 0. (x – y)2 0, which is known to be true.

What is wrong?

Lecture 2 – Feb 19, 2003

Proving Existential Statements Proofs of existential statements are also

called existence proofs. Two types of existence proofs

Constructive Construct the object. Prove that it has the necessary properties.

Non-constructive Argue indirectly that the object must exist.

Example: Constructive Proof Theorem: Given a segment AB, there is a

midpoint M of AB. Proof:

A B

M

C

Justification Argue by SAS that triangles ACM and BCM

are congruent and that AM = MB.

A B

M

C

Example: Constructive Proof Theorem: The equation

x2 – 7y2 = 1.

has a solution in positive integers. Proof:

Let x = 8 and y = 3. Then 82 – 732 = 64 – 63 = 1.

Example: Constructive Proof Theorem: The equation

x2 – 67y2 = 1.

has a solution in positive integers. Proof: ?

Example: Non-Constructive Proof Theorem: There exists x R such that

x5 – 3x + 1 = 0. Proof:

Let f(x) = x5 – 3x + 1. f(1) = –1 < 0 and f(2) = 27 > 0. f(x) is a continuous function. By the Intermediate Value Theorem, there

exists x [1, 2] such that f(x) = 0.

Disproving Universal Statements Construct an instance for which the

statement is false. Also called proof by counterexample.

Example: Proof by Counterexample Disprove the conjecture (Fermat): All

integers of the form 22n + 1, for n 1, are prime.

(Dis)proof: Let n = 5. 225 + 1 = 4294967297. 4294967297 = 6416700417.

Example: Proof by Counterexample Disprove the statement: If a function is

continuous at a point, then it is differentiable at that point.

(Dis)proof: Let f(x) = |x| and consider the point x = 0. f(x) is continuous at 0. f(x) is not differentiable at 0.

Disproving Existential Statements These can be among the most difficult of all

proofs. Famous examples

There is no formula “in radicals” for the general solution of a 5th degree polynomial.

There is no solution in positive integers of the equation

xn + yn = zn.

Example: Disproving an Existential Statement Theorem: There is no solution in integers to the

equation

x2 – y2 = 101010 + 2. Proof:

A perfect square divided by 4 has remainder 0 or 1. Therefore, x2 – y2 divided by 4 has remainder 0, 1, or 3. However, 101010 + 2 divided by 4 has remainder 2. Therefore, x2 – y2 101010 + 2 for any integers x and y.

Section 3.2

Direct Proof and Counterexample II: Rational Numbers

Rational Numbers A rational number is a number that equals

the quotient of two integers. Let Q denote the set of rational numbers. An irrational number is a number that is

not rational. We will assume, for the time being, that

there exist irrational numbers.

Direct Proof Theorem: The sum of two rational numbers

is rational. Proof:

Let r = a/b and s = c/d be rational. Then r + s = (ad + bc)/bd, which is rational.

Proof by Counterexample Disprove: The sum of two irrationals is

irrational. Counterexample:

Let α be irrational. Then –α is irrational. α + (–α) = 0, which is rational.

Direct Proof Theorem: Between every two distinct

rationals, there is a rational. Proof:

Let r, s Q. Assume that r < s. Let t = (r + s)/2. Then t Q. We must show that r < t < s.

Proof continued Given: r < s. Add r: 2r < r + s. Divide by 2: r < (r + s)/2 = t. Given: r < s. Add s: r + s < 2s. Divide by 2: t = (r + s)/2 < s. Therefore, r < t < s.

Other Theorems Theorem: Between every two distinct

irrationals there is a rational. Proof: ? Theorem: Between every two distinct

irrationals there is an irrational. Proof: ?

An Interesting Question Why are the last two theorems so hard to

prove? Because they involve “negative”

hypotheses and “negative” conclusions.

Positive and Negative Statements A positive statement asserts the existence of

a number. A negative statement asserts the

nonexistence of a number. It is much easier to use a positive

hypothesis than a negative hypothesis. It is much easier to prove a positive

conclusion than a negative conclusion.

Positive and Negative Statements “r is rational” is a positive statement.

It asserts the existence of integers a and b such that r = a/b.

“α is irrational” is a negative statement. It asserts the nonexistence of integers a and b

such that α = a/b. Is there a “positive” characterization of

irrational numbers?

Section 3.3

Direct Proof and Counterexample III: Divisibility

Divisibility Definition: An integer a divides an integer

b if a 0 and there exists an integer c such that

ac = b. Write a | b to indicate that a divides b. Divisibility is a positive property.

Units Definition: An integer u is a unit if u | 1. This is a positive property.

Why? The only units are 1 and –1.

Composite Numbers Definition: An integer n is composite if

there exist non-units a and b such that

n = ab. A composite number factors in a non-trivial

way. Is this a positive property?

What about the non-units?

Prime Numbers Definition: An integer p is prime if p is not

a unit and p is not composite. A prime number factors only in a trivial

way. This is a negative property. Prime numbers: 2, 3, 5, 7, 11, …

Example: Direct Proof Theorem: If u and v are units, then uv is a unit. Proof:

Let u and v be units. There exist integers r and s such that ur = 1 and vs = 1. Therefore, (ur)(vs) = 1. Rearrange: (uv)(rs) = 1. Therefore, uv is a unit.

Example: Direct Proof Theorem: Let a and b be integers. If a | b and

b | a, then a/b and b/a are units. Proof:

Let a and b be integers. Suppose a | b and b | a. There exist integers c and d such that ac = b and bd = a. Therefore, acd = bd = a. Therefore, cd = 1. Thus, c and d are units.

Corollary: If a | b and b | a, then a = b or a = –b.

Section 3.4

Direct Proof and Counterexample IV: Division into Cases and the

Quotient-Remainder Theorem

The Quotient-Remainder Theorem Theorem: Let n and d be integers, d 0.

Then there exist unique integers q and r such that

n = qd + r

and

0 r < d. q is the quotient and r is the remainder.

Example: Proof by Cases Theorem: For any integer n, n3 – n is a

multiple of 6. Proof:

Divide n by 6 to get q and r: n = 6q + r, where 0 r < 6.

Substitute: n3 – n = (6q + r)3 – (6q + r). Expand and rearrange:

n3 – n = 6(36q3 + 18q2r + 3qr2 – q) + (r3 – r).

Proof continued Therefore, 6 | (n3 – n) if and only if 6 | (r3 – r). Consider the 6 possible cases:

Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60. Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60. Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61. Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64. Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610. Case 6: r = 5. r3 – r = 53 – 5 = 120 = 620.

Proof continued In every case, 6 | (r3 – r).

Therefore, 6 | (r3 – r) in general. Therefore, 6 | (n3 – n) for all integers n.

Section 3.5

Direct Proof and Counterexample V: Floor and Ceiling

The Floor Function Let x be a real number. The floor of x,

denoted x, is the integer n such that

n x < n + 1. If x is an integer, then x = x. If x is not an integer, then x is the first integer

such that x < x.

The Ceiling Function The ceiling of x, denoted x, is the integer n

such that

n – 1 < x n. If x is an integer, then x = x. If x is not an integer, then x is the first integer

such that x > x.

Example: Direct Proof Theorem: Let x and y be real numbers.

Thenx + y x + y < x + y + 1.

Proof (1st inequality): By definition, x x and y y. Therefore, x + y x + y.

Proof (2nd inequality): By definition, x + y < x + y + 1.

Exercise: The Ceiling Function Theorem: Let x and y be real numbers.

Then

x + y – 1 < x + y x + y. Proof: Exercise

Questions Is – –x = x true for all real numbers x?

Proof: ? Is x – 1 < x x true for all real numbers x?

Proof: ? Is 2x + 2y = x + y + x + y true for all

real numbers x and y? Proof: ?

An Interesting Theorem Theorem: Let x be a positive real number. Then x

is irrational if and only if the two sequences

1 + x, 2 + 2x, 3 + 3x, …and

1 + 1/x, 2 + 2/x, 3 + 3/x, …together contain every positive integer exactly once.

Proof: ?

Section 3.6

Indirect Argument: Contradiction and Contraposition

Form of Proof by Contraposition Theorem: p q. This is logically equivalent to q p. Outline of the proof of the theorem:

Assume q. Prove p. Conclude that p q.

This is a direct proof of the contrapositive.

Benefit of Proof by Contraposition If p and q are negative statements, then p

and q are positive statements. We may be able to give a direct proof that

q p.

Example: Proof by Contraposition Theorem: The sum of a rational and an

irrational is irrational. Restate the theorem: Let r be a rational

number and let α be a number. If α is irrational, then r + α is irrational.

Restate again: Let r be a rational number and let α be a number. If r + α is rational, then α is rational.

The Proof Proof:

Let r be rational and α be a number. Suppose that r + α is rational. Let s = r + α. Then α = s – r, which is rational. Therefore, if r + α is rational, then α is rational. It follows that if α is irrational, then r + α is

irrational.

Example: Proof by Contraposition Theorem: If u is a unit and p is prime, then

up is prime. Restatement: Let u be unit and p be an

integer. If p is a prime, then up is a prime. 2nd Restatement: Let u be unit and p be an

integer. If up is not a prime, then p is not a prime.

Proof continued Proof:

Let u be a unit and p an integer. There is an integer v such that uv = 1. Suppose up is not prime. Two possibilities:

up is a unit. up is composite.

Proof continued (Case 1) Case 1: up is a unit.

Then (up)v is a unit. However, (up)v = (uv)p = p. Therefore, p is a unit. Therefore, p is not a prime.

Proof continued (Case 2) Case 2: up is composite.

There exist non-units b and c such that up = bc. Then p = (uv)p = (up)v = (bc)v = (bv)c. bv and c are non-units. Therefore, p is composite. Therefore, p is not a prime.

Proof concluded In both cases p is not a prime. Therefore p is not a prime in general. Therefore, if p is prime, then up is prime.

Form of Proof by Contradiction Theorem: p q. Outline of the proof of the theorem :

Assume (p q). This is equivalent to assuming p q. Derive a contradiction, i.e., conclude r r for

some statement r. Conclude that p q.

Benefit of Proof by Contradiction The statement r may be any statement

whatsoever because any contradiction

r r

will suffice.

Contradiction vs. Contraposition Sometimes a proof by contradiction “becomes” a

proof by contraposition. Here is how it happens.

Assume (p q), i.e., p q. Prove p. Cite the contradiction p p. Conclude that p q.

Is this proof by contradiction or by contraposition? Proof by contraposition is preferred.

Useful Fact Theorem: An integer p is prime if and only

if, for all integers a and b, if p | ab, then p | a or p | b.

In symbols, p is prime if and only ifa, b Z, (p | ab p | a p | b)

This is a positive characterization of primes. It may allow a direct proof rather than a

proof by contradiction or contraposition.

Direct Proof Theorem: If u is a unit and p is prime, then

up is prime. Proof:

Let u be a unit and p a prime. There is an integer v such that uv = 1. Let a and b be integers and suppose that up | ab. There exists an integer c such that upc = ab. Therefore, p | ab.

Section 3.7

Two Classical Theorems

Classical Theorem #1 Theorem: 2 is irrational. Proof (Euclid):

Suppose 2 is rational. There exist integers a and b such that 2 = a/b. (WOLOG) Assume that a and b are relatively

prime. Square: 2b2 = a2. Therefore, 2 | a2 and so 2 | a.

Proof concluded Substitute 2c for a: 2b2 = 4c2. Simplify: b2 = 2c2. Therefore, 2 | b2 and so 2 | b. This contradicts the assumption that a and b are

relatively prime. Therefore, 2 is irrational.

Classical Theorem #2 Theorem: The set of prime numbers is infinite. Proof:

Suppose there are only finitely many primes. Let {p1, …,pn} be a complete list of the primes. Let k = (p1 … pn) + 1. k 2, yet pi does not divide k for any i. This is a contradiction. Therefore, there are infinitely many primes.

Euclid’s proof.

Example: Constructive Existence Proof Theorem: Between any two distinct

irrationals there is a rational and an irrational.

Proof: Let α and β be irrational numbers with α < β. Then β – α > 0. Choose an integer n such that n(β – α) > 1. Then 1/n < β – α.

Proof continued Let m = nβ – 1. Then

m < nβ m + 1. Then m/n < β and nβ – 1 m. Then α < β – 1/n = (nβ – 1)/n m/n. Therefore, α < m/n < β.

Proof concluded Choose an integer k such that

k(β – m/n) > 2. Divide by k:

β – m/n > 2/k. Then β > m/n + 2/k. Therefore,

α < m/n < m/n + 2/k < β.

lecture 3 – february 17, 2003. chapter 3 elementary number theory and methods of proof

Documents

proof of theorem proof

equation x n

direct proof slide

methods of proof slide

instance slide

integers x

c slide

introduction slide