Lecture 3 – February 17, 2003

Elementary Number Theory and Methods of Proof Chapter 3 Elementary Number Theory and Methods of Proof

Direct Proof and Counterexample I: Introduction 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).

Example: Direct Proof 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.

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 – 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: C A B M

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 – 732 = 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 = 6416700417.

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.

Direct Proof and Counterexample II: Rational Numbers 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. 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. Add s: r + s < 2s. Divide by 2: t = (r + s)/2 < s. Therefore, r < t < s.

Lecture 3 – Feb 20, 2003

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?

Direct Proof and Counterexample III: Divisibility 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.

Example: Direct Proof Theorem: Let a, b, c be integers. If a | b and b | a + c, then a | c. Proof: Let a, b, and c be integers. Suppose a | b and b | a + c. There exist integers d and e such that ad = b and be = a + c. Substitute: (ad)e = a + c. Rearrange: a(de – 1) = c. Therefore, a | c.

Lecture 12 – Feb 20, 2003

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 = 60. Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60. Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61. Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64. Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610. Case 6: r = 5. r3 – r = 53 – 5 = 120 = 620.

Proof continued Therefore, 6 | (r3 – r) in general. In every case, 6 | (r3 – r). Therefore, 6 | (r3 – r) in general. Therefore, 6 | (n3 – n) for all integers n.

Direct Proof and Counterexample V: Floor and Ceiling 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.

x + y  x + y < x + y + 1. Example: Direct Proof Theorem: Let x and y be real numbers. Then x + 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? Is 2x + 2y = x + y + x + y true for all real numbers x and y?

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: ?

Lecture 13 – Feb 20, 2003

Indirect Argument: Contradiction and Contraposition 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.

Lecture 14 – Feb 24, 2003

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 if a, 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. 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.

Proof concluded Thus, p | a or p | b, since p is prime. Case 1: p | a. Then there exists an integer d such that pd = a. Then (up)(dv) = (uv)(pd) = a. Therefore, up | a. Case 2: p | b. Similar to Case 1. Therefore, up | a or up | b. Therefore, up is prime.

Two Classical Theorems 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.

α < m/n < m/n + 2/k < 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.