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# mt3sol - Homework 3 Math 245 Fall 2010 Due Friday October 1...

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Homework 3, Math 245, Fall 2010 Due Friday, October 1 in class. The solution of each exercise should be at most one page long. If you can, try to write your solutions in LaTex. Each question is worth 2 points. 1. The Fibonacci numbers ( F n ) n 1 are defined as follows: F 1 = F 2 = 1 and F n = F n - 1 + F n - 2 for every n 2. Prove that F n = 1+ 5 2 n - 1 - 5 2 n 5 for every n 1. Proof. We will use strong induction on n to prove that F n = 1+ 5 2 n - 1 - 5 2 n 5 . Base Case: n = 1. In this case, F 1 = 1 and 1+ 5 2 1 - 1 - 5 2 1 5 = 2 5 2 5 = 1. n = 2. In this case, F 2 = 1 and F 2 = 1+ 5 2 2 - 1 - 5 2 2 5 = 6+2 5 4 - 6 - 2 5 4 5 = 4 5 4 5 = 1 . Thus, the base case is true. Induction Step: Assume that F i = 1+ 5 2 i - 1 - 5 2 i 5 for every i , 1 i k . We will prove that F k +1 = 1+ 5 2 k +1 - 1 - 5 2 k +1 5 (1)

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Using the recurrence relation of the Fibonacci numbers and the induction hypothesis, we get F k +1 = F k + F k - 1 = 1+ 5 2 k - 1 - 5 2 k 5 + 1+ 5 2 k - 1 - 1 - 5 2 k - 1 5 = 1+ 5 2 k + 1+ 5 2 k - 1 5 - 1 - 5 2 k + 1 - 5 2 k - 1 5 = 1+ 5 2 k - 1 1+ 5 2 + 1 5 - 1 - 5 2 k - 1 1 - 5 2 + 1 5 = 1+ 5 2 k - 1 · 3+ 5 2 5 - 1 - 5 2 k - 1 · 3 - 5 2 5 = 1+ 5 2 k - 1 · 6+2 5 4 5 - 1 - 5 2 k - 1 · 6 - 2 5 4 5 = 1+ 5 2 k - 1 · 1+ 5 2 2 5 - 1 - 5 2 k - 1 · 1 - 5 2 2 5 = 1+ 5 2 k +1 5 - 1 - 5 2 k +1 5 . This finishes our proof. 2. Let f : A B and g : B C be two functions. Show that if g f is injective, then f is injective. Show that if g f is surjective, then g is surjective. Proof. First, we show that if g f is injective, then f injective. Let a, b A such that f ( a ) = f ( b ). Applying g to both sides we obtain g ( f ( a )) = g ( f ( b )) which means ( g f )( a ) = ( g f )( b ). As g f is injective, this implies a = b and shows f is injective. Secondly, we show that if g f is surjective, then g is surjective. Let c C . As g f is surjective, it means there exists a A such that ( g f )( a ) = c .
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