By Thomas Keilen

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Q 0 q/d , so r r 0 is divisible by d . But jr r 0 j Ä maxfr; r 0 g < d , so the only possibility is r r 0 D 0. q 0 q/d D 0, so q 0 q D 0. ■ We have shown the existence of a prime factorization of any natural number, but we have not shown that the prime factorization is unique. This is a more subtle issue, which is addressed in the following discussion. The key idea is that the greatest common divisor of two integers can be computed without knowing their prime factorizations. 8. A natural number ˛ is the greatest common divisor of nonzero integers m and n if (a) ˛ divides m and n and (b) whenever ˇ 2 N divides m and n, then ˇ also divides ˛.

M; n/ and nr is a common divisor of m and n. 9, nr is the greatest common divisor of m and n. m; n/. 11. Find the greatest common divisor of 1734282 and 452376. 1734282; 452376/. We can find the coefficients s; t such that 18 D s 1734282 C t 452376: The sequence of quotients q1 ; q2 ; : : :Ä; q6 in the algorithm is 3; 1; 5; 72; 19; 3. 0 1 The qk determine matrices Qk D . The coefficients s; t com1 qk prise the first column of Q D Q1 Q2 Q6 . 12. m; n/ D 1. ✐ ✐ ✐ ✐ ✐ ✐ “bookmt” — 2006/8/8 — 12:58 — page 32 — #44 ✐ 32 ✐ 1.

Q 0 q/d D 0, so q 0 q D 0. ■ We have shown the existence of a prime factorization of any natural number, but we have not shown that the prime factorization is unique. This is a more subtle issue, which is addressed in the following discussion. The key idea is that the greatest common divisor of two integers can be computed without knowing their prime factorizations. 8. A natural number ˛ is the greatest common divisor of nonzero integers m and n if (a) ˛ divides m and n and (b) whenever ˇ 2 N divides m and n, then ˇ also divides ˛.