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Unformatted text preview: 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 + m 2. Then if m 6 = m , 2 = n n m m Q , which is false. Therefore m = m n = n . This proves that, nb 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...
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This document was uploaded on 03/27/2010.
 Spring '09

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