Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Math 347 - Fall 2015 Midterm 3 practice solutions 1. Prove from the definition of a convergent sequence that if an → L and bn → M , then an + b n → L + M . Solution. This is theorem 14.5(a) in the book. 2. (a) Prove that every infinite subset of a countable set is countable. (b) Prove that every set that contains an uncountable set is uncountable. Solution. This is problem 13.7 on homework 6. 3. (a) State the definition of a Cauchy sequence and the Cauchy convergence criterion. (b) State the definition of a subsequence and the Bolzano–Weierstrass Theorem. Solution. These are Definition 14.12, Theorem 14.19, and Theorem 14.17 in the book. 4. Let a, b ∈ N. (a) State the definition of gcd(a, b). (b) Prove that the set of integer combinations of a and b equals the set of integer multiples of gcd(a, b). Solution. a) gcd(a, b) is the largest natural number that divides both a and b. b) This is Theorem 6.12 in the book. 5. Prove that a bounded sequence need not be Cauchy by providing a counterexample and a formal -style argument showing that your sequence is not a Cauchy sequence. Solution. Consider an = (−1)n and take = 1/2 and set m = n + 1. Then for all N , if n, m ≥ N we have |an − am | = |an − an+1 | = |2| ≥ 1/2 = , so the sequence is not Cauchy. 6. The set of irrational numbers is R \ Q. Prove that the irrational numbers are uncountable. Solution. We argue by contradiction, so assume R \ Q is countable, and let f : N → R \ Q be a bijection. Since Q is countable, there is also a bijection g : N → Q. We use these bijections to define a new bijection, h : N → R, by f (n/2) n even h(n) = . g((n + 1)/2) n odd Since h has an inverse h−1 : R → N defined by 2f −1 (x) x∈R\Q −1 h (x) = −1 2g (x) − 1 x∈Q h is indeed a bijection as claimed. But since R is uncountable, there cannot exist a bijection h : N → R, which gives a contradiction to R \ Q being countable. Hence, R \ Q is uncountable. 1 7. (a) State the comparison test for series. Solution. See Proposition 14.29 in the book. (b) Prove by induction that 3n − 1 ≥ 2n when n ≥ 1. Solution. For a base case, we have 31 − 1 = 2 ≥ 21 . For the inductive step, assume that 3n − 1 ≥ 2n up to some n. Then 3n+1 − 1 = 3(3n − 1) + 2 > 3(3n − 1) > 2(3n − 1) ≥ 2 · 2n = 2n+1 where the last greater than or equal to sign uses the inductive hypothesis. This completes the induction. ∞ X 1 (c) Prove that converges. n 3 − 1 n=1 Solution. By the comparison test, we have 0≤ 1 1 ≤ 3n − 1 2n P∞ P 1 and since the series ∞ n=1 n=1 2n converges (since it is a geometric series) the series also converges by the comparison test. 1 3n −1 8. For the following questions, decide whether the statement is true or false and provide a brief proof or counterexample justifying your choice. P (a) If lim an does not exist, then ∞ n=1 an is divergent. n→∞ P Solution. TRUE. The contrapositive of the statement reads “If ∞ n=1 an is convergent, then limn→∞ an exists” is true, since this limit must exist and equal zero. (b) If lim an = 1 and lim bn does not exist, then lim an bn does not exist. n→∞ n→∞ n→∞ Solution. TRUE. We argue by contradiction. Suppose limn→∞ an bn = L. Then since 1/an → 1/1 = 1, we have that L = lim an bn lim (1/an ) = lim (an bn /an ) = lim bn , n→∞ n→∞ n→∞ n→∞ but since limn→∞ bn does not exist, this is a contradiction. Note: It would be wrong to start out by writing lim an bn = lim an lim bn n→∞ n→∞ n→∞ since this formula only holds with both limits on the right exist. (c) If lim (an+k − an ) = 0 for every k ∈ N, then {an } converges. n→∞ √ Solution. FALSE. The sequence an = n provides a counterexample since √ √ lim ( n + k − n) = lim √ n→∞ n→∞ for every k ∈ N. 2 k √ =0 n+k+ n (d) If ∞ X a2k diverges, then ∞ X ak diverges. k=1 k=1 √ P 2 Solution. FALSE. A counterexample isPak = (−1)k / k. Then the series ∞ k=1 ak is the harmonic series (which diverges) while ∞ k=1 ak converges. 9. (a) Compute the gcd of 78 and 90 using the Euclidean algorithm. Solution. We did this one in review: the Euclidean algorithm for this pair is gcd(78, 90) = = = = gcd(90, 78) = gcd(90 − 78, 78) = gcd(12, 78) gcd(78, 12) = gcd(78 − 6 · 12, 12) = gcd(6, 12) gcd(12, 6) = gcd(12 − 2 · 6, 6) = gcd(0, 6) 6. (b) Compute the gcd of 78 and 90 using prime factorization. (HINT: 78 = 6 · 13.) Solution. We have 90 = 9 · 10 = 2 · 32 · 5 = 6 · 15 and 78 = 6 · 13. Since 13 and 15 are relatively prime, we have gcd(90, 78) = 6. 10. (a) Carefully explain how an infinite decimal expansion of a real number defines a Cauchy sequence whose limit is the real number given by the decimal expansion. Solution. An infinite decimal expansion of a real number x ∈ [0, 1] defines a sequence k X an xk = 10n n=1 with xk → x. This sequence is Cauchy since for all m ≥ n ≥ N we have |xm − xn | < 1 1 ≤ 10n−1 10N −1 (b) Give a careful proof (using the definition of a convergent sequence) that .999 · · · = 1. Solution. The infinite decimal .9̄ is the limit of the sequence k X 9 . xk = 10n n=1 We have 1 − xk = n ≥ N, 1 . 10k Hence, for any given > 0, there exists an N such that for all 1 1 ≤ N < n 10 10 and hence the limit of the sequence xk is 1. This shows .9̄ = 1. |1 − xn | = 3