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Unformatted text preview: 1 For all k ≥ 1 Prove that For all n ∈ Z + , gcd( u n +1 , u n ) = 1. Proof. We proceed by induction on n . Base case (n=1) Then we want to show that ( u 1 , u 2 ) = 1. By defnition oF the ±ibonacci sequence, u 1 = u 2 = 1. Moreover, For any d ∈ Z + , ( d, d ) = d since d  d . ThereFore, (1 , 1) = 1 and we are done. Inductive step Suppose that k ∈ Z + and ( u k +1 , u k ) = 1. We want to show that ( u k +2 , u k +1 ) = 1. u k +2 = u k +1 + u k = u k +1 (1) + u k . ThereFore, by Theorem 16.1.2 (with a = u k +2 , b = u k +1 and r = u k ), ( u k +2 , u k +1 ) = ( u k +1 , u k ) . Moreover, by the inductive hypothesis, this is 1. ±...
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 Spring '06
 Knutson
 Math, Division, Integers, Natural number, Greatest common divisor, Euclidean algorithm, Euclidean domain

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