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30 Slides--Induction II

# 30 Slides--Induction II - CS103 HO#30 Slides-Induction II...

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CS103 HO#30 Slides--Induction II April 26, 2010 1 The Principle of Mathematical Induction A proof by mathematical induction that a proposition P n is true for every positive integer n consists of two steps: BASE CASE: Show that P 1 is true. INDUCTIVE STEP: Assume that P k is true for an arbitrarily chosen positive integer k and show that under that assumption, P k+1 must be true. From these two steps we conclude (by the principle of mathematical induction) that for all positive integers n, P n is true. Note that in the inductive step, you are proving a conditional: IF P k for some arbitrary k, THEN P k+1 . P 1 k P k . . . . . . P k+1 n P n Here is a Formal version Peano's Axioms (1889) • The equality relation is reflexive, symmetric, and transitive, and the natural numbers are closed under equality. • There is a natural number 0. • Every natural number has a successor, denoted by S(a). • There is no natural number whose successor is 0, i.e., x (S(x) 0). • Two natural numbers with the same successor are themselves equal, i.e., x y(S(x) = S(y) x = y) • If a property is possessed by 0 and if the successor of every natural number possessing the property also possesses it, then it is possessed by every number, i.e., [Q(0) ∧ ∀ x(Q(x) Q(S(x))] → ∀ xQ(x) Bad Induction Show that for any positive n, n 3 – n + 1 is a multiple of 3. P n : n 3 – n + 1 is a multiple of 3 We will show that if we assume P k is true, then P k+1 is true (k + 1) 3 – (k + 1) + 1 = k 3 + 3k 2 + 3k + 1 – k = k 3 – k + 1 + 3(k 2 + k) Divisible by 3 by inductive hypothesis Obviously divisible by 3 So P k+1 is true and P n is true for all n. What's wrong? Bad Induction Show that for any positive n, n 3 – n + 1 is a multiple of 3. P n : n 3 – n + 1 is a multiple of 3 We will show that if we assume P k is true, then P k+1 is true (k + 1) 3 – (k + 1) + 1 = k 3 + 3k 2 + 3k + 1 – k = k 3 – k + 1 + 3(k 2 + k) Divisible by 3 by inductive hypothesis Obviously divisible by 3 So P k+1 is true and P n is true for all n. What's wrong? There is no base case. Not only is P 1 not true, P n is false for all n . The Well-Ordering Property Every non-empty set of positive integers has a least element. From this axiom, we can prove that the Principle of Mathematical Induction is correct. The Well-Ordering Property Every non-empty set of positive integers has a least element. From this axiom, we can prove that the Principle of Mathematical Induction is correct. Suppose that for some P n we show the base case P 1 and the inductive step k 1 (P k P k+1 ). In order to derive a contradiction, suppose that it is not the case that nP n where n 1. Then n(¬P n ).

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