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Unformatted text preview: Section 10 – Natural Numbers and Induction WellOrdering Property of N Our study of the natural numbers N begins with the wellordering property and the principle of mathematical induction. MAT 300 Awtrey MATHEMATICS AND STATISTICS 1 / 15 Section 10 – Natural Numbers and Induction WellOrdering Property of N Our study of the natural numbers N begins with the wellordering property and the principle of mathematical induction. Axiom ( WellOrdering Property of N ) If S is a nonempty subset of N , then there exists an element m 2 S such that m k for all k 2 S. MAT 300 Awtrey MATHEMATICS AND STATISTICS 1 / 15 Section 10 – Natural Numbers and Induction WellOrdering Property of N Our study of the natural numbers N begins with the wellordering property and the principle of mathematical induction. Axiom ( WellOrdering Property of N ) If S is a nonempty subset of N , then there exists an element m 2 S such that m k for all k 2 S. Theorem ( Principle of Mathematical Induction ) Let P ( n ) be a statement that is either true or false for each n 2 N . Then P ( n ) is true for all n 2 N provided that (a) P ( 1 ) is true ( the base case ) (b) and for each k 2 N , if P ( k ) is true, then P ( k + 1 ) is true ( the induction step ) MAT 300 Awtrey MATHEMATICS AND STATISTICS 1 / 15 Section 10 – Natural Numbers and Induction Proof of Mathematical Induction We will prove this by contradiction: suppose P ( 1 ) is true and if P ( k ) is true then P ( k + 1 ) is true for all k 2 N . And suppose there exists an n 2 N such that P ( n ) is false. MAT 300 Awtrey MATHEMATICS AND STATISTICS 2 / 15 Section 10 – Natural Numbers and Induction Proof of Mathematical Induction We will prove this by contradiction: suppose P ( 1 ) is true and if P ( k ) is true then P ( k + 1 ) is true for all k 2 N . And suppose there exists an n 2 N such that P ( n ) is false. Consider the set S = f n 2 N : P ( n ) is false g MAT 300 Awtrey MATHEMATICS AND STATISTICS 2 / 15 Section 10 – Natural Numbers and Induction Proof of Mathematical Induction We will prove this by contradiction: suppose P ( 1 ) is true and if P ( k ) is true then P ( k + 1 ) is true for all k 2 N . And suppose there exists an n 2 N such that P ( n ) is false. Consider the set S = f n 2 N : P ( n ) is false g Thus S is nonempty and the WellOrdering Property guarantees the existence of an element m 2 S that is the least element of S . MAT 300 Awtrey MATHEMATICS AND STATISTICS 2 / 15 Section 10 – Natural Numbers and Induction Proof of Mathematical Induction We will prove this by contradiction: suppose P ( 1 ) is true and if P ( k ) is true then P ( k + 1 ) is true for all k 2 N . And suppose there exists an n 2 N such that P ( n ) is false. Consider the set S = f n 2 N : P ( n ) is false g Thus S is nonempty and the WellOrdering Property guarantees the existence of an element m 2 S that is the least element of S . Since P ( 1 ) is true, 1 = 2 S so m > 1....
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This note was uploaded on 02/22/2011 for the course MAT 300 taught by Professor Thieme during the Spring '07 term at ASU.
 Spring '07
 thieme
 Math, Statistics, Natural Numbers, Mathematical Induction

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