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Published byFernanda Hanford Modified over 2 years ago

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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}. 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.

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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.

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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

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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

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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 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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