Problem_Set_04_Solutions

Problem_Set_04_Solutions - ArsDigita University Month 2:...

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ArsDigita University Month 2: Discrete Mathematics - Professor Shai Simonson Problem Set 4 – Induction and Recurrence Equations Thanks to Jeffrey Radcliffe and Joe Rizzo for many of the solutions. Pasted together and modified by Dimitri Kountourogiannis 1. What’s wrong with the following proofs by induction? (a) All binary strings are identical. The proof is by induction on the size of the string. For n= 0 all binary strings are empty and therefore identical. Let X = b n b n - 1 ...b 1 b 0 be an arbitrary binary string of length n+ 1. Let Y = b n b n - 1 ...b 1 and Z = b n - 1 ...b 1 b 0 . Since both Y and Z are strings of length less than n+ 1, by induction they are identical. Since the two strings overlap, X must also be identical to each of them. ANSWER This is not a subtly flawed proof. It is a terribly flawed proof. The problem is with the induction step. It is never true that the pasting together of two identical strings is identical to the strings. Furthermore, the n = 1 case fails for another reason: A string of length 1 is not the pasting together of two strings of length 0. (b) Any amount of change greater than or equal to twenty can be gotten with a combination of five cent and seven cent coins. The proof is by induction on the amount of change. For twenty cents use four five-cent coins. Let n > 20 be the amount of change. Assume that n = 7 x + 5 y for some non-negative integers x and y. For any n > 20, either x > 1, or y > 3. If x > 1, then since 3(5) – 2(7) = 1, n+ 1 = 5( y +3)+7( x –2). If y > 3, then since 3(7) – 4(5) = 1, n+ 1 = 7( x +3)+5( y –4). In either case, we showed that n+ 1 = 7 u + 5 v where u and v are non-negative integers. ANSWER For any n > 20, either x > 1 or y > 3 is false. For example, if n = 21, then 21 = 0 · 5 + 3 · 7. 2. Prove by induction that: (a) The n th Fibonacci number, F n equals G n = 1 5 " ( 1 + 5 2 ) n - ( 1 + 5 2 ) n # where F 0 = 0 and F 1 = 1. ANSWER Base Case: These hold since G 0 = 1 5 " ( 1 + 5 2 ) 0 - ( 1 + 5 2 ) 0 # = 1 5 (1 - 1) = 0 and G 1 = 1 5 " ( 1 + 5 2 ) 1 - ( 1 + 5 2 ) 1 # = 1 5 " 1 + 5 2 - 1 - 5 2 # = 1 Induction Step: Suppose that G n = F n and G n - 1 = F n - 1 We would like to show G n +1 = F n +1 Let a = ( 1 2 + 5 2 ) Let b = ( 1 2 - 5 2 ) We will use the fact that a , b are roots of r 2 - r - 1 = 0 and hence a 2 = a + 1 b 2 = b + 1 1
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Starting with G n +1 , G n +1 = 1 5 ( a n +1 - b n +1 ) = 1 5 ( a 2 a n - 1 - b 2 b n - 1 ) = 1 5 (( a + 1) a n - 1 - ( b + 1) b n - 1 ) = 1 5 ( a n - b n ) + 1 5 ( a n - 1 - b n - 1 )) = F n + F n - 1 = F n +1 So the induction step is proven. QED. (b) The sum of the geometric series 1 + a + a 2 + ...+ a n equals (1– a n +1 )/(1– a ), where a does not equal one.
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Problem_Set_04_Solutions - ArsDigita University Month 2:...

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