klarfeld-sec-1.6-2.1 - Jennifer Klarfeld Homework Section...

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Proof by Contradiction [ ( m ) ( n ) ( 12 m + 15 n = 1 ) ] Suppose [ ( m ) ( n ) ( 12 m + 15 n = 1 ) ] which is equivalent to ( m ) ( n ) ( 12 m + 15 n = 1 ) . Jennifer Klarfeld Homework Section 1.6-2.1 Section 1.6 1. Prove that: (b) there exist integers m and n such that 15 m + 12 n = 3 . Let m = 1 n =− 1. Then 15 ( 1 )+ 12 (− 1 )= 15 12 = 3. Therefore,there exist integers m nsuchthat 15 m + 12 n = 3, namelym = 1 n =− 1. (d) there do not exist integers m and n such that 12 m + 15 n = 1 . Suppose thereexist integers m nsuchthat 12 m + 15 n = 1. Then 3 ( 4 m + 5 n )= 1 4 m + 5 n = 1 3 . Because m nareintegers, 4 m + 5 nisaninteger . But ,based onour assumption, 4 m + 5 n = 1 3 ,whichis not aninteger . Thus,we havea contradiction. Therefore,there doesnot exist integersm nsuchthat 12 m + 15 n = 1. 2. Prove that for all integers a , b , and c : 1
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Jennifer Klarfeld Homework Section 1.6-2.1 (a)if a divides b 1 and a divides c 1 , then a divides bc-1. Suppose a b 1 a c 1. Thenb 1 = ak c 1 = aj for someintegers k j.
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