45 Slides--Induction II

45 Slides--Induction II - CS103A HO #45 Induction II...

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Unformatted text preview: CS103A HO #45 Induction II 2/25/08 1 Peano's Axioms There is a number 0. Every number has a successor, denoted by S(a). There is no number whose successor is 0, i.e., x (S(x) 0). Two 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 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) 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 Fitch version Show that for any positive integer n, n 5 n is divisible by 5. Proof by Mathematical Induction. P(n): n 5 n is divisible by 5 BASE CASE: P(1) asserts that 1 5 1 is divisible by 5. 1 5 1 = 0, and since 0 is divisible by 5, P(1) is true. INDUCTIVE STEP: Assume for some positive integer k, P(k): k 5 k is divisible by 5 Show P(k + 1): (k + 1) 5 (k + 1) is divisible by 5 Show that for any positive integer n, n 5 n is divisible by 5. Proof by Mathematical Induction. P(n): n 5 n is divisible by 5 BASE CASE: P(1) asserts that 1 5 1 is divisible by 5. 1 5 1 = 0, and since 0 is divisible by 5, P(1) is true. INDUCTIVE STEP: Assume for some positive integer k, P(k): k 5 k is divisible by 5 Show P(k + 1): (k + 1) 5 (k + 1) is divisible by 5 Proof of the inductive step: (k + 1) 5 (k+1)=k 5 + 5 k 4 + 10 k 3 + 10 k 2 + 5k + 1 k 1 = (k 5 k)+5(k 4 + 2 k 3 + 2 k 2 + k) Since k 5 k is divisible by 5 by the inductive hypothesis, the RHS is divisible by 5 and P(k + 1) is true. Thus n 5 n is divisible by 5 for any n > 0 by the principle of mathematical induction. Bad Induction Show that for any positive n, n 3 n + 1 is a multiple of 3. 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) CS103A HO #45 Induction II 2/25/08 2 Bad Induction Show that for any positive n, n 3 n + 1 is a multiple of 3....
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This note was uploaded on 10/01/2011 for the course CS 103A taught by Professor Plummer,r during the Winter '07 term at Stanford.

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45 Slides--Induction II - CS103A HO #45 Induction II...

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