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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.
- Spring '08