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Why is the remainder a mod ba \bmod b smaller than bb in the Euclidean step?

By the definition of division with remainder, for any integers aa and bb (b>0b>0), there exist unique integers qq and rr such that a=bq+ra = bq + r and 0≤r<b0 \le r < b. This constraint on rr ensures that the remainder is always strictly smaller than the divisor bb, which is crucial for the algorithm's termination.

Conditions

  • bb is a positive integer.
  • aa is a nonnegative integer.
  • The operation is a mod ba \bmod b.

Reasoning, step by step

  1. Recall the Division Algorithm theorem.
  2. It states a=bq+ra = bq + r with 0≤r<b0 \le r < b.
  3. The inequality r<br < b is part of the definition of the remainder.
  4. Therefore, the remainder is always smaller than the divisor.
  5. This property allows the Euclidean algorithm to reduce the size of the numbers in each step.

Example

The script states: 'For b=aq+r with 0≤r<a0\le r<a...' (Note: The script uses b=aq+rb=aq+r with 0≤r<a0 \le r < a, implying aa is the divisor in that specific notation context, but the principle is that the remainder is less than the divisor).

Common misconceptions

  • Thinking that the remainder can be equal to the divisor.
  • Believing that the remainder can be negative.
  • Confusing the divisor and the dividend in the inequality condition.

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