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# section10 - Section 10 Natural Numbers and Induction...

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Section 10 – Natural Numbers and Induction Well-Ordering Property of N Our study of the natural numbers N begins with the well-ordering property and the principle of mathematical induction. MAT 300 Awtrey MATHEMATICS AND STATISTICS 1 / 15

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Section 10 – Natural Numbers and Induction Well-Ordering Property of N Our study of the natural numbers N begins with the well-ordering property and the principle of mathematical induction. Axiom ( Well-Ordering 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 Well-Ordering Property of N Our study of the natural numbers N begins with the well-ordering property and the principle of mathematical induction. Axiom ( Well-Ordering 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

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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

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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 Well-Ordering 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 Well-Ordering 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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