The well ordering property
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THE WELL ORDERING PROPERTY. Definition: Let B be a set of integers. An integer m is called a least element of B if m is an element of B, and for every x in B, m x. Example: 3 is a least element of the set {4,3,5,11}.

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THE WELL ORDERING PROPERTY

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The well ordering property

THE WELL ORDERING PROPERTY

Definition: Let B be a set of integers. An integer m is called a least element ofB if m is an element of B, and for every x in B, mx.

Example: 3 is a least element of the set {4,3,5,11}.

Example: Let A be the set of all positive odd integers. Then 1 is a least element of A.

Example: Let U be the set of all odd integers. Then U has no least element. The proof is left as an informal exercise.


Well ordering

WELL ORDERING

The Well Ordering Property: Let B be a non empty set of integers such that there exists an integer b such that every element x of B satisfies

b < x. Then there exists a least element of B.

Therefore, in particular: If B is a non empty set of non negative integers then there exists a least element in B.


Well ordering1

WELL ORDERING

An Application of the Well Ordering Property: A Proof of the Existence Part of the Division Algorithm:

For every integer a and positive integer d there exist integers q and r such that 0r<d and a = qd + r.


Well ordering2

WELL ORDERING

Proof: Let A be the set of all values a – kd for integers k, such that a – kd is non negative. We show that A is not empty: If a is non negative, let k=0. a-kd=a, and a is non negative. If a is

negative, let k=a. a-kd=a-ad=a(1-d). (1-d) is not

positive (since d is positive), and a is negative,

so the product a(1-d) is non negative.

By the well ordering property A has a least element. Let that least element be r = a – qd. So a = qd + r and 0r. We claim that r<d. This claim can be proved by contradiction.


Well ordering3

WELL ORDERING

Suppose that dr. Note that

a = qd + r = qd + d + (r-d) = (q+1)d + r’

where r’ = (r-d)  0. Also r’ = r-d < r, because d > 0. So

a – (q+1)d = r’ < r = a – qd, and

a – (q+1)d = r’  0. So a – (q+1)d is in A, and a – (q+1)d < a – qd – contradicting the fact that a – qd is the least element of A.


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