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hw5sol

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

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Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 5 (due February 23, 2009) § 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 · 2 / 2) 2 = 1 2 = 1, so P (1) holds. c) The inductive hypothesis is P ( n ), that 1 3 + 2 3 + · · · + 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 + · · · + 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, we’ve shown P (1). In part e, we’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. § 4.1, #6: When n = 1, then 1 · 1! = 1 and (1 + 1)! - 1 = 2 - 1 = 1, so the statement holds. 1

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For the inductive step, suppose that the statement holds for n = k , then, for n = k + 1, the left hand side is, 1 · 1! + · · · + k · k ! + ( k + 1) · ( k + 1)! = ( k + 1)! - 1 + ( k + 1) · ( k + 1)! = ( k + 2) · ( 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 . § 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 · 2 + 1 2 · 3 + · · · 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 · 2 + 1 2 · 3 + · · · + 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. § 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 . § 4.1, #50: For n = 1, the statement is that for a set of 1 + 1 = 2 positive integers, none exceeding 2 · 1 = 2, there is at least one integer that divides another. But, the only possible such set is { 1 , 2 } , and 1 divides 2. Therefore, the statement is true for n = 1. For the inductive step, we assume that for every set of n +1 integers at most 2 n , there is an integer in the set dividing another in it. Let S be a set of n +2 integers which are at most 2( n +1). Either S contains neither 2 n + 1 nor 2 n + 2, S contains one of them, or S contains both. In the first case, for an arbitrary element k in S , S -{ k } has n +1 elements at most 2 n , so by the inductive hypothesis, one of them divides another. In the second case, S -{ 2 n +1 , 2 n +2 } has the same property, so again, by the inductive hypothesis, one integer in S divides another.

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

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