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Unformatted text preview: 16 Solutions for Exercises: Elementary Number Theory and Methods of Proof Chapter 4: Elementary Number Theory and Methods of Proof Section 4.1 3. a . Yes, because 4 rs = 2 (2 rs ) and 2 rs is an integer since r and s are integers and products of integers are integers. b . Yes, because 6 r + 4 s 2 + 3 = 2(3 r + 2 s 2 + 1) + 1 and 3 r + 2 s 2 + 1 is an integer since r and s are integers and products and sums of integers are integers. c . Yes, because r 2 + 2 rs + s 2 = ( r + s ) 2 and r + s is an integer that is greater than or equal to 2 since both r and s are positive integers and thus each is greater than or equal to 1. 6. For example, let a = 1 and b = 0. Then a + b = 1 = 1 and also a + b = 1 + 0 = 1 . Hence for these values of a and b, a + b = a + b. In fact, if a is any nonzero integer and b = 0, then a + b = a + 0 = a = a + 0 = a + 0 = a + b. 12. Counterexample : Let n = 5 . Then n 1 2 = 5 1 2 = 4 2 = 2 , which is not odd. 15. According to the order of operations for real numbers, a n = ( a n ). The following table shows that the property is true for some values of a and n and false for other values. a n a n ( a ) n Does a n = ( a ) n ? 2 2 = 0 = 0 ( 0) 2 = 0 2 = 0 Yes 3 2 3 2 = (3 2 ) = 9 ( 3) 2 = ( 3)( 3) = 9 No 2 3 ( 2) 3 = (( 2) 3 ) = ( 8) = 8 ( ( 2)) 3 = 2 3 = 8 Yes 3 2 ( 3) 2 = (( 3) 2 ) = 9 ( ( 3)) 3 = 3 2 = 9 No 18. 1 2 1 + 11 = 11, which is prime. 2 2 2 + 11 = 13, which is prime. 3 2 3 + 11 = 17, which is prime. 4 2 4 + 11 = 23, which is prime. 5 2 5 + 11 = 31, which is prime. 6 2 6 + 11 = 41, which is prime. 7 2 7 + 11 = 53, which is prime. 8 2 8 + 11 = 67, which is prime. 9 2 9 + 11 = 83, which is prime. 10 2 10 + 11 = 101, which is prime. Section 4.1 17 21. If a real number is greater than 1, then its square it greater than itself. Start of Proof : Suppose x is any [particular but arbitrarily chosen] real number such that x > 1. Conclusion to be shown: x 2 > x . 27. Proof 1 : Suppose m and n are any [particular but arbitrarily chosen] odd integers. [We must show that m + n is even.] By definition of odd, there exist integers r and s such that m = 2 r + 1 and n = 2 s + 1. Then m + n = (2 r + 1) + (2 s + 1) by substitution = 2 r + 2 s + 2 = 2( r + s + 1) by algebra. Let u = r + s + 1. Then u is an integer because r , s , and 1 are integers and a sum of integers is an integer. Hence m + n = 2 u , where u is an integer, and so by definition of even, m + n is even [as was to be shown]. Proof 2 : Suppose m and n are any [particular but arbitrarily chosen] odd integers. [We must show that m + n is even.] By definition of odd, there exist integers r and s such that m = 2 r + 1 and n = 2 s + 1. Then m + n = (2 r + 1) + (2 s + 1) by substitution = 2 r + 2 s + 2 = 2( r + s + 1) by algebra....
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 Spring '11
 Shirley
 Number Theory, Integers

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