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PS1-Solutions

# PS1-Solutions - MAT 337 Problem Set 1 Sample Solutions Ian...

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MAT 337 Problem Set 1 Sample Solutions Ian Zwiers - February 10, 2009 This document is for demonstration purposes; it is not a marking key. 2.4 A(f) Compute the limit lim n →∞ n 2 + n - n ; then use = 10 - 6 to find integer N that satisfies the defition. Multiplying by the conjugate, n 2 + n - n = n n 2 + n + n = 1 1+ 1 n +1 . Thus, 1 2 = 1 1 + 1 = 1 q 1 + lim n →∞ 1 n + 1 = 1 lim n →∞ q 1 + 1 n + 1 = lim n →∞ p n 2 + n - n. Since 1 n > 0, p n 2 + n - n - 1 2 = 1 2 - 1 q 1 + 1 n + 1 < 1 2 - 1 ( 1 + 1 n ) + 1 = 1 2 (2 n + 1) . Note that 1 2(2 n +1) < 10 - 6 whenever 10 6 4 - 1 2 < n . Thus, for = 10 6 , N = 10 6 4 satisfies the definition. 2.5 (I) Show the set S = n + m 2 : m, n Z is dense in R . Suppose that n + m 2 = n 0 + m 0 2. Then if m 6 = m 0 , 2 = n - n 0 m 0 - m Q , which is false. Therefore m = m 0 n = n 0 . This proves that, n -b m 2 c + m 2 m Z + } ⊂ S [0 , 1] has an infinite number of elements. Let > 0 be arbitrary. We may divide [0 , 1] into N = d 1 e subintervals I 1 , I 2 , . . . I N , each with length no more than . We have proven S [0 , 1] is infinite, so there are at least N + 1 elements of S in [0 ,

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PS1-Solutions - MAT 337 Problem Set 1 Sample Solutions Ian...

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