College Algebra Exam Review 21

# College Algebra Exam Review 21 - we have .n r ;0/ D .m;n/Q...

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1.6. DIVISIBILITY IN THE INTEGERS 31 This process must stop after no more than n steps with some remainder n r C 1 D 0 . Then we have the following system of relations: m D q 1 n C n 1 n D q 2 n 1 C n 2 ::: n k ± 2 D q k n k ± 1 C n k ::: n r ± 1 D q r C 1 n r : Proposition 1.6.10. The natural number n r is the greatest common divisor of m and n , and furthermore n r 2 I.m;n/: Proof. Write m D n ± 1 and n D n 0 . It is useful to consider each step of the algorithm as producing a new pair .n k ± 1 ;n k / from the previous pair .n k ± 2 ;n k ± 1 / by a linear transformation: .n k ± 1 ;n k / D .n k ± 2 ;n k ± 1 / ± 0 1 1 ± q k ² The matrix Q k D ± 0 1 1 ± q k ² is invertible with inverse Q ± 1 k D ± q k 1 1 0 ² . Set Q D Q 1 Q 2 ²²² Q r C 1 ; then both Q and Q ± 1 have integer entries and
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Unformatted text preview: we have .n r ;0/ D .m;n/Q and .m;n/ D .n r ;0/Q 1 : Therefore n r D sm C tn , where s t is the rst column of Q , and .m;n/ D .n r a;n r b/ , where a b is the rst row of Q 1 . It follows that n r 2 I.m;n/ and n r is a common divisor of m and n . By the observation follow-ing Proposition 1.6.9 , n r is the greatest common divisor of m and n . n We denote the greatest common divisor by g.c.d. .m;n/ ....
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