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The Euclidean algorithm is an efficient method for computing the gcd of two positive integers. Given two positive integers a and b (with a 鈮� b), the algorithm is as follows: 1. Divide a by b and get the remainder r鈧� (a = b*q鈧� + r鈧�, where q鈧� is the quotient). 2. If r鈧� = 0, then gcd(a, b) = b. 3. Else, replace a with b and b with r鈧�, and repeat the process until the remainder is 0.

Let's apply the algorithm to get the sequence of remainders: 1. a = b * q鈧� + r鈧� 2. b = r鈧� * q鈧� + r鈧� 3. r鈧� = r鈧� * q鈧� + r鈧� ... n. r鈧欌倠鈧� = r鈧欌倠鈧� * q鈧� + r鈧� (with r鈧� 鈮� 0) n+1. r鈧欌倠鈧� = r鈧� * q鈧欌倞鈧� (with remainder 0) The Euclidean algorithm will stop at the (n+1)th step since the remainder r鈧� is zero.

We would use mathematical induction on the remainders to prove this: Base case, n = 1: We need to prove gcd(a, b) = gcd(b, r鈧�). According to Step 2, we have a = b * q鈧� + r鈧�. Since a is a multiple of gcd(a, b) and a - b * q鈧� = r鈧�, then r鈧� must also be a multiple of gcd(a, b). So gcd(a, b) must divide both b and r鈧�, and hence gcd(a, b) 鈮� gcd(b, r鈧�). Furthermore, since gcd(b, r鈧�) divides both b and r鈧�, it must also divide a, so gcd(b, r鈧�) 鈮� gcd(a, b). Combining these inequalities, we have gcd(a, b) = gcd(b, r鈧�). Induction step: Suppose the statement is true for n = k, i.e., gcd(r鈧栤倠鈧�, r鈧�) = gcd(a, b). We must now prove it for n = k + 1, i.e., gcd(r鈧�, r鈧栤倞鈧�) = gcd(a, b). For n = k + 1, we have r鈧� = r鈧栤倞鈧� * q鈧栤倞鈧� + r鈧栤倞鈧� with remainder 0. So, gcd(r鈧栤倠鈧�, r鈧�) = gcd(r鈧�, r鈧栤倞鈧�). Since gcd(r鈧栤倠鈧�, r鈧�) = gcd(a, b) (from our induction hypothesis), we get gcd(a, b) = gcd(r鈧�, r鈧栤倞鈧�). Hence, by the principle of mathematical induction, we have proven that gcd(a, b) = gcd(r鈧欌倠鈧�, r鈧�) for all n 鈮� 1.