These slides are a summary of the Well-Ordering Principle. Video explains these slides is available in this link https://youtu.be/EkleZiBtYyk Reference books for these slides are A Transition to Advanced Mathematics 8th Edition, by Douglas Smith, Maurice Eggen, Richard St. Andre. ISBN-13: 978-1285463261, published by Cengage Learning (August 6, 2014). https://www.cengagebrain.co.uk/shop/isbn/9781285463261 and Discrete Mathematics with Applications, 3nd Edition, (1995) By Susanna S. Epp, ISBN13: 9780534359454, published by Thomson-Brooks/Cole Publishing Company.

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