Lecture Week 4

# Discrete Mathematics and Its Applications with MathZone

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UCI ICS/Math 6D, Fall 2007 4-Induction-1 Induction Let P(n) be a predicate (“propositional function”) regarding integers n≥n 0 for some n 0 (typically n 0 =0 or 1). Normal” Induction Normal” Induction : If you prove two things: 1) Basis step : P(n 0 ) is true and 2) Inductive step : For all k≥n 0 : If P(k) true then P(k+1) true then it follows that P(n) is true for all n≥n 0 . Example: Show that P(n) = “1+2+...+n=n(n+1)/2” is true for all n≥1. 1) Basis (for n 0 =1): “1=1·(1+1)/2” is true. 1) Inductive step: Assuming P(k) [that is, assuming 1+2+...+k=k(k+1) ] prove P(k+1) [that is, prove 1+2+...+k+(k+1)=(k+1)(k+2)/2 ] Ref: The inductive step is a proof of an implication P(k) => P(k+1)

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UCI ICS/Math 6D, Fall 2007 4-Induction-2 Induction Examples For n≥0, n 2 <4 n . P(n)=“n 2 <4 n Basis step: P(0)=“0<4 0 =1” is true Inductive step: Assuming k 2 <4 k , prove (k+1) 2 <4 k+1 . The sum of the first n odd positive integers is n 2 Basis step: “1=1 2 ” is true Inductive step: Assuming 1+3+5+...+(2k-1)=k 2 , prove that 1+3+5+...+(2k-1)+(2k+1)= (k+1) 2 A set with n elements has 2 n subsets (|P(S)|=2 |s| ) Basis step: |P( ) |=|{ }|=1=2 0
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