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Answers for “为什么余数总是小于 $b$?”

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In the Euclidean step gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b)=\gcd(b,a\bmod b), the remainder a mod ba\bmod b becomes the new second argument of the pair. It replaces the original dividend aa in the next iteration.

Conditions: aa is a nonnegative integer.; bb is a positive integer.; The operation is the Euclidean step gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b)=\gcd(b,a\bmod b).