109_sp2011_ho_division_sol

109_sp2011_ho_division_sol - 1 For all k ≥ 1 Prove that...

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109 Spring 2011 - Division Algorithm Exercise. Find the greatest common divisor of 2047 and 1633 and ±nd integers m and n such that 460 = 2047 m + 1633 n. We use the Euclidean algorithm: 2047 = 1633(1) + 414 1633 = 414(3) + 391 414 = 391(1) + 23 391 = 23(17) + 0 . Therefore (2047 , 1633) = (1633 , 414) = (414 , 391) = (391 , 23) = 23 . Reading the computation backwards gives us 23 as an integral linear combination of 2047 and 1633: 23 = 414 391(1) = 414 [1633 414(3)] = 414(4) + 1633( 1) = [2047 1633](4) + 1633( 1) = 2047(4) + 1633( 5) . Notice that 460 = 23 · 20. Therefore, 460 = 20 ± 2047(4) + 1633( 5) ² = 2047(80) + 1633( 100) . Hence, m =80and n = 100.
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Exercise (IV.7) . Recall the defnition oF the ±ibonacci sequence (5.4.2): u 1 =1 u 2 =1 u k +1 = u k + u k -
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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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This note was uploaded on 08/02/2011 for the course MATH 109 taught by Professor Knutson during the Spring '06 term at UCSD.

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109_sp2011_ho_division_sol - 1 For all k ≥ 1 Prove that...

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