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Unformatted text preview: Solutions to Homework 5 Section 4.1 4. (a) Plugging in n = 1 we have that P (1) is the statement 1 3 = [1 (1+1) / 2] 2 . (b) Both sides of P (1) shown in part (a) equal 1. (c) The inductive hypothesis is the statement that 1 3 + 2 3 + + k 3 = k ( k + 1) 2 2 . (d) For the inductive step, we want to show for each k 1 that P ( k ) implies P ( k +1) . In other words, we want to show that assuming the inductive hypothesis we can prove [1 3 + 2 3 + + k 3 ] + ( k + 1) 3 = ( k + 1)( k + 2) 2 2 . (e) Replacing the quantity in brackets on the lefthand side of part (d) by what it equals by virtue of the inductive hypothesis, we have k ( k + 1) 2 2 + ( k + 1) 3 = ( k + 1) 2 k 2 4 + k + 1 = ( k + 1)( k + 2) 2 2 as desired. (f) We have completed both the basis step and the inductive step, so by the principle of mathematical induction, the statement is true for every positive integer n. 20. The basis step is n = 7 , and indeed 3 7 < 7! , since 2187 < 5040 . Assume the statement for k. Then 3 k +1 = 3 3 k < ( k + 1) 3 k < ( k + 1) k ! = ( k + 1)! , the statement for k + 1 . 1 38. The basis step is trivial, as usual: A 1 B 1 implies that S 1 j =1 A j S 1 j =1 B j because the union of one set is itself. Assume the inductive hypothesis that if A j B j for j = 1 , 2 ,...,k , then S k j =1 A j S k j =1 B j . We want to show that if A j B j for j = 1 , 2 ,...,k + 1 , then S k +1 j =1 A j S k +1 j =1 B j . To show that one set is a subset of another we show that an arbitrary element of the first set must be an element of the second set. So let x S k +1 j =1 A j = S k j =1 A j A k +1 . Either x S k j =1 A j or x A k +1 . In the first case we know by the inductive hypothesis that....
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This note was uploaded on 06/07/2011 for the course COT 3100 taught by Professor Staff during the Summer '08 term at University of Florida.
 Summer '08
 Staff

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