We want to prove that if a and b are both odd, and a b, then gcd(a, b) = gcd((a b)/2, b). If a and b are both odd and a b, then gcd(a, b) = gcd(a b, b) because, from the regular Euclidean algorithm, gcd(a, b) begins with a = qb + r, 0 r < b and gcd(a b, b) begins with a b = (q 1)b + r, 0 r < b. The next step in either case is to divide b by r, so the two final answers will be the same. Finish this proof by showing that gcd(a b, b) = gcd((a b)/2, b).

# We want to prove that if a and b are both odd, and a b, then gcd(a, b) = gcd((a b)/2

ISBN: 9781429215107
256

## Solution for problem 25 Chapter 2.3

Mathematical Structures for Computer Science | 7th Edition

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We want to prove that if a and b are both odd, and a b, then gcd(a, b) = gcd((a b)/2