hw5sol

# hw5sol - Math 55 Discrete Mathematics UC Berkeley Spring...

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Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2010 Solutions to Homework #5 (due on 3-3-2010) ยง 4.1, #4: a) The statement P (1) is that 1 3 = (1(1 + 1) / 2) 2 . b) The left hand side is 1 and the right hand side is (1 ยท 2 / 2) 2 = 1 2 = 1, so P (1) holds. c) The inductive hypothesis is P ( n ), that 1 3 + 2 3 + ยทยทยท + n 3 = ( n ( n + 1) / 2) 2 . d) To prove the inductive step, you need to prove that if the inductive hypothesis P ( n ) is true, then P ( n + 1) is true. e) We assume the inductive hypothesis. Then, 1 3 + 2 3 + ยทยทยท + n 3 + ( n + 1) 3 = n ( n + 1) 2 2 + ( n + 1) 3 = n 2 ( n + 1) 2 + 4( n + 1) 3 4 = ( n 2 + 4 n + 4)( n + 1) 2 4 = ( n + 2) 2 ( n + 1) 2 4 = ( n + 1)( n + 2) 2 2 , which is the statement P ( k + 1). f) In part b, weโve shown P (1). In part e, weโve shown that P (1) implies P (2), which implies P (3), which implies P (4), and so on. For any n , this chain of implications means that P ( n ) holds. ยง 4.1, #6: When n = 1, then 1 ยท 1! = 1 and (1 + 1)!- 1 = 2- 1 = 1, so the statement holds. 1 For the inductive step, suppose that the statement holds for n = k , then, for n = k + 1, the left hand side is, 1 ยท 1! + ยทยทยท + k ยท k ! + ( k + 1) ยท ( k + 1)! = ( k + 1)!- 1 + ( k + 1) ยท ( k + 1)! = ( k + 2) ยท ( k + 1)!- 1 = ( k + 2)!- 1 , which is the desired formula for n = k + 1. By induction, the equation holds for all positive integers n . ยง 4.1, #10 a) For small values, we have, 1 2 = 1 2 1 2 + 1 6 = 4 6 = 2 3 1 2 + 1 6 + 1 12 = 9 12 = 3 4 From this, I would guess that 1 1 ยท 2 + 1 2 ยท 3 + ยทยทยท 1 n ( n + 1) = n n + 1 b) The base case, when n = 1 was checked in part a. By induction, assume that the formula holds for n = k . Then the left hand for n = k + 1 is 1 1 ยท 2 + 1 2 ยท 3 + ยทยทยท + 1 k ( k + 1) + 1 ( k + 1)( k + 2) = k k + 1 + 1 ( k + 1)( k + 2) = k ( k + 2) + 1 ( k + 1)( k + 2) = k 2 + 2 k + 1 ( k + 1)( k + 2) = ( k + 1) 2 ( k + 1)( k + 2) = k + 1 k + 2 which is the equation from part a for n = k + 1. ยง 4.1, #33: For n = 0, n 5- n = 0- 0 = 0, which is divisible by 5. 2 By induction, assume that n 5- n is divisible by 5 for n = k . Then, for n = k + 1, ( k + 1) 5- ( k + 1) = k 5 + 5 k 4 + 10 k 3 + 10 k 2 + 5 k + 1- k- 1 = ( k 5- k ) + 5( k 4 + 2 k 3 + 2 k 2 + k ) By the inductive hypothesis, k 5- k is divisible by 5, and the second term is divisible by 5 because it is 5 times an integer, so ( k +1) 5- ( k +1) is divisible by 5. Therefore, by induction n 5- n is divisible by 5 for all non-negative integers n . ยง 4.1, #50: For n = 1, the statement is that for a set of 1+1 = 2 positive integers, none exceeding 2 ยท 1 = 2, there is at least one integer that divides another. But, the only possible such set is { 1 , 2 } , and 1 divides 2....
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## This note was uploaded on 04/05/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

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hw5sol - Math 55 Discrete Mathematics UC Berkeley Spring...

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