Homework 9 Solutions

# Homework 9 Solutions - a,b ] /b is an integer. Hence a = ab...

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MATH 74: HOMEWORK 9 COMMENTS Book Problem # 13 The least common multiple of two nonzero integers a and b is the unique positive integer m such that (i) m is a common multiple, i.e., a | m and b | m , (ii) m is the smallest common multiple: if n | a and n | b then m n. We denote the least common multiple of a and b by [ a,b ]. Proposition 0.1. Let m = [ a,b ] as above. If a | n and b | n then m | n . Proof. By the division theorem there exist unique integers q and r such that n = mq + r 0 r < m. If r = 0 then m | n and we are done. So suppose that r > 0. Since a | n and a | m we know there exist integers t 1 and q 1 such that n = at 1 and m = aq 1 . Hence n = mq + r = at 1 = aq 1 q + r = r = a ( t 1 - q 1 q ) . Hence a | r . Similarly, we can prove that b | r . By the second property of the deﬁnition of [ a,b ] we see that m r . But this contradicts the fact (furnished by the division theorem above) that m > r . This means that the assumption that r > 0 is wrong, and so m | n , as claimed. ± By Proposition 0.1 with n = ab , it follows that [ a,b ] | ab . Hence ab/ [ a,b ] is an integer. Let k = ab/ [ a,b ] to avoid clutter. We claim that k | a and k | b . We’ll show that k | a ; the other claim is similar. By deﬁnition of m we know that b | [ a,b ], hence [

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Unformatted text preview: a,b ] /b is an integer. Hence a = ab [ a,b ] · [ a,b ] b = k · [ a,b ] b and so k | a . By deﬁnition of greatest common divisor, it follows that k ≤ ( a,b ), i.e., (1) ab [ a,b ] ≤ ( a,b ) . Finally, we claim that p := ab/ ( a,b ) is a common multiple of a and b , i.e., a | p and b | p . From the deﬁnition of greatest common divisor, we know that ( a,b ) | b , whence b/ ( a,b ) is an integer. Then p = ab ( a,b ) = a · b ( a,b ) Date : April 10th, 2008. 1 shows that a | p . The claim that b | p is proved similarly. It follows from the deﬁnition of least common multiple that (2) ab ( a,b ) ≥ [ a,b ] Combining ( 1 ) and ( 2 ) we obtain ( a,b )[ a,b ] = ab, as desired. To compute [612 , 696], ﬁrst use the Euclidean algorithm to show that (612 , 696) = 12. Now [612 , 696] = 612 × 696 (612 , 696) = 612 × 696 12 = 51 × 696 = 35 , 496 . 2...
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## This note was uploaded on 06/15/2009 for the course MATH 74 taught by Professor Courtney during the Spring '07 term at Berkeley.

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Homework 9 Solutions - a,b ] /b is an integer. Hence a = ab...

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