Well-Ordering Principle
- 1. Well-Ordering Principle Dr. Yassir Dinar Spring 2020 Dr. Yassir Dinar ell-Ordering Principle Spring 2020 1 / 5
- 2. Well-Ordering Principle (WOP) Well-Ordering Principle (WOP). Every nonempty set of integers whose elements are greater than some real number, has a smallest element. Thus, if S be a subset of Z satisfies 1 S = ∅ 2 (∃m ∈ R)(∀t ∈ S)(m < t) Then (∃M ∈ S)(∀t ∈ S)(M ≤ t). Example 3.1 Which of the following sets of integers has a smallest element 1 The set of all prime numbers greater than 8. 2 The set of positive real numbers. Solution. 1 It has a smallest element by WOP. 2 It has no smallest element. Dr. Yassir Dinar ell-Ordering Principle Spring 2020 2 / 5
- 3. Examples WOP (with Direct Proof) WOP for N. Every nonempty subset of N has a smallest element. Example 3.2 Let a and b be nonzero integers. Then there is a smallest positive linear combination of a and b. Solution. Let a and b be nonzero integers. Let S be the set of all positive linear combination on a and b. By definition elements of S are greater than 0. Also S = ∅ since if a > 0, then a = a(1) + b(0) ∈ S and if a < 0, then −a = a(−1) + b(0) ∈ S. By the WOP, the set S has a smallest element r. Then, by definition, r is the smallest positive linear combination of a and b. Dr. Yassir Dinar ell-Ordering Principle Spring 2020 3 / 5
- 4. Examples WOP (with Proof by Contradiction) Example 3.3 Every natural number greater than 1 is prime or is a product of primes. Solution. Assume there is a natural number greater than 1 which is not prime and is not a product of primes. Then the set T of such numbers is not empty. By WOP, T contains a smallest number m. Since m is not prime then m is a composite. Thus m = sr for some natural numbers 1 < s < m and 1 < r < m. By the choice of m, r and s are not in T. Hence each of s and r either is prime or is a product of primes. Thus m = sr is a product of primes, a contradiction. Therefore, every natural number greater than 1 is prime or is a product of primes. Dr. Yassir Dinar ell-Ordering Principle Spring 2020 4 / 5
- 5. Examples WOP (With Direct Proof) Theorem 3.4 (The Division Algorithm) For all integers a and b with a > 0, there exist unique integers q and r such that b = aq + r and 0 ≤ r < a. Solution. Existence:Let S = {b − ak : k ∈ Z and b − ak ≥ 0}. If b ≥ 0, then b − (0)a ∈ S and if b < 0, then b − 2ba = b(1 − 2a) ∈ S. Thus S = ∅. Hence, by WOP, there is a smallest element r ∈ S. Then there exits integer q such that r = b − aq. Thus b = aq + r. Verify conditions: By assumptions r ≥ 0. Suppose a ≤ r. Then 0 ≤ r − a = b − a(q + 1) ≤ r. Hence r − a ∈ S and r is not the smallest element in S, a contradiction. Hence r < a. Uniqueness:Assume b = aq1 + r1 = aq2 + r2 for some integers q1, q2, r1, r2 with 0 ≤ r1, r2 < a. Assume r1 ≤ r2.Then r2 − r1 = a(q1 − q2) and a|(r2 − r1). Since 0 ≤ r2 − r1 < a, r2 − r1 = 0. But then q1 − q2 = 0. Thus r1 = r2 and q1 = q2. Dr. Yassir Dinar ell-Ordering Principle Spring 2020 5 / 5