Adv Alegbra HW Solutions 25

Adv Alegbra HW Solutions 25 - 25 Therefore, 1 is a linear...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
25 Therefore, 1 is a linear combination of rr 0 and m ; as 1 is obviously the smallest positive linear combination, it must be their gcd. 1.59 Let a , b and d be integers. If d = sa + tb , where s and t are integers, f nd in f nitely many pairs of integers ( s k , t k ) with d = s k a + t k b . Solution. If d = sa + tb , and if we de f ne s k = s + kb and t k = t ka , then d = s k a + t k b for all k . 1.60 If a and b are relatively prime and if each divides an integer n , prove that their product ab also divides n . Solution. Assume that ( a , b ) = 1 and n = ak = b ` . By Corollary 1.40, b | ak implies b | k . Thus, k = bk 0 and so n = ak = abk 0 . 1.61 Prove, for any (possibly negative) integers a and b , that ( b , a ) = ( b a , a ) . Solution. If c is a common divisor of a and b , then c | a and c | b ; hence, c | b a , and c is a common divisor of a and b a . This does not yet show that the two gcd s are equal. However, if c 0 is a common divisor of a and b a , then
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online