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Section 1.2

1(d) -162=(17)(-10) + 8, so q=-10 and r=8. Remember that tex2html_wrap_inline43 .

2(b) 39214=(871)(45)+19 so r=19=39214-(871)(45).

6. S'pose the three consecutive integers are a, a+1 and a+2. If a is divisible by 3, then we're done. Assume 3 does not divide a. Then use the division algorithm to write

displaymath63

with r=1 or r=2. If r=1 then

displaymath71

(add 2 to both sides of the other equation). Since 3 divides the right hand side, 3 divides a+2. Similarly, if r=2 then 3 divides a+1. Therefore, 3 must divide one of a, a+1 or a+2.

9(b). Use the Euclidean Algorithm:

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displaymath93

displaymath95

displaymath97

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Therefore, gcd(41, 25)=1 (you could also do this using the prime factorization of 41 and 25). Working backwards, we have

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displaymath105

displaymath107

displaymath109

displaymath111

44. We show the representation exists by strong induction on n. If n=0, take tex2html_wrap_inline117 . If n>0, write tex2html_wrap_inline121 with tex2html_wrap_inline123 using the division algorithm. then

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and tex2html_wrap_inline127 because q<0 implies tex2html_wrap_inline131 , contrary to the hypotheses. Hence by induction, tex2html_wrap_inline133 , with tex2html_wrap_inline135 for all i. The representation of n now follows because tex2html_wrap_inline121 . To prove uniqueness, suppose tex2html_wrap_inline143 is another such representation Then tex2html_wrap_inline145 and tex2html_wrap_inline147 are each the remainder when n is divided by b, so tex2html_wrap_inline153 . Now tex2html_wrap_inline155 , so tex2html_wrap_inline157 follows the same way. The continues to show tex2html_wrap_inline159 for all tex2html_wrap_inline161 .



Kristine Bauer
Wed Feb 4 13:45:43 MST 2004