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Answers for “为什么余数严格小于除数?”

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The Euclidean algorithm reduces the problem size by replacing the larger number with the remainder of the division of the two numbers. Since the remainder rr satisfies 0≤r<a0 \le r < a (where aa is the divisor), the new number is strictly smaller than the previous larger number, ensuring progress towards the base case.

Conditions: The current pair is (a,b)(a,b) with b≥a>0b \ge a > 0.; The step computes b=aq+rb = aq + r with 0≤r<a0 \le r < a.; The new pair is (r,a)(r,a) or (a,r)(a,r) depending on implementation, but effectively the max value decreases.