key1fa00 - COT3100C-01, Fall 2000 S. Lang Solution Key to...

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COT3100C-01, Fall 2000 S. Lang Solution Key to Assignment #1 (40 pts.) 9/14/2000 1. (16 pts.) Recall the following definitions and theorems about integers: Definition . An integer a is even if a = 2 b for some integer b . (That is, there exists an integer b such that a = 2 b .) Definition . An integer a is odd if a = 2 b + 1 for some integer b . (That is, there exists an integer b such that a = 2 b + 1.) Definition . An integer a is a divisor of integer b , denoted a | b , if a 0 and there exists integer c such that b = ac . Theorem. Each integer is either even or odd (but not both). Theorem. The sum of two odd integers is even. Theorem. If the product of two integers is even, then at least one of them is even. Theorem. If a | b and b | c , then a | c . Use these definitions and theorems (and other appropriate laws) to answer each of the following questions, where all variables refer to integers: (a) If a + b + c is odd, then either all 3 numbers a , b , and c are odd or exactly one of them is odd. We prove the following contrapositive statement, using the indirect proof method: If (all 3 numbers a , b , c , are even) or (exactly one of the 3 numbers is even), then the sum a + b + c is even --- (1). We consider two cases: (Case One) All 3 numbers a , b , c , are even. That is, a = 2 p , b = 2 q , and c = 2 r , for some integers p , q , and r . Thus, a + b + c = 2 p + 2 q + 2 r = 2( p + q + r ) --- (2). Since the factor ( p + q + r ) in (2) is an integer by the closure property of integer addition, so (2) implies a + b + c is even. (Case Two) Exactly one of the 3 numbers a , b , c , is even. We first assume a is even, both b and c are odd. Thus, by the definition of even and odd integers, a = 2 p , b = 2 q + 1, and c = 2 r + 1. Adding them up, we get a + b + c = 2 p + (2 q + 1) + (2 r + 1) = 2( p + q + r + 1) --- (3). Since the factor ( p + q + r + 1) in (3) is an integer by the closure property of integer addition, so (3) implies
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key1fa00 - COT3100C-01, Fall 2000 S. Lang Solution Key to...

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