Count the final division producing zero as an iteration, and let the total be . The terminal product alone gives no useful bound. Instead apply (i) to each nonterminal product, which is a positive integer and hence at least .
If , the penultimate product satisfiesAs is an integer, this impliesFor , the same bound holds since implies . In particular the Euclidean algorithm uses divisions, in big O notation. The convention that includes the final zero-producing division explains the one-step offset in the argument.
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