Hw4 - We prove a^n = 1 for all n(Here ^ represents exponentiation so"a^n" is the variable a to the n power Base case a^0 = 1 is true by

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EE 369 Homework 4 ============================================================================== REMINDER: Collaboration must end before writing begins (on each problem separately). See the initial course information handout. ============================================================================== Please read sections 2.1 and 2.2 in the Gersting textbook The problems from section 2.1 and 2.2 apply to your course outcome 2 score. Solve the following problems. 5 points per problem, except 8 points for number 9 (sec 1.4) and 12 points for number 25 (sec 1.4). From Section 2.1: problems 2, 3, 13, 21, 26, 40, 50, 59 From Section 2.2: problems 10, 16, 40, 43, 64, 70 E1. Argue that the "strong induction principle" (the "second induction principle" in your book) follows from the principle of well ordering. E2. What is wrong with the following proof by strong induction:
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Unformatted text preview: We prove a^n = 1 for all n. (Here, ^ represents exponentiation, so "a^n" is the variable a to the n power) Base case: a^0 = 1 is true by definition. Inductive case: assume a^m = 1 for all m such that 0 <= m <= k. We must then use the assumption to show a^(k+1) = 1. Observe that a^(k+1) = (a^k)(a^k) / (a^(k-1)). But by our induction hypothesis, both a^k and a^(k-1) are 1. So, a^(k+1) = (1)(1)/(1) which is 1, as desired. E3. Show that if a_1, a_2, . .., a_n are n distinct real numbers, then exactly n-1 multiplications are used to compute the product of these n numbers no matter how parentheses are inserted into their product. (Hint: use the 2nd principle of mathematical induction and consider the last multiplication done)....
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This note was uploaded on 09/24/2009 for the course ECE 369 taught by Professor Bagchi during the Spring '08 term at Purdue University-West Lafayette.

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