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Math 142A Homework Assignment 3 Selected Solutions 2.2.4 Show that the set of irrational numbers fails to be closed. Solution Consider the sequence { πn }. Observe that nπ is an irrational number for every index n; for, if r = πn were rational, then n r would be a product of a natural number and a rational number equal to the irrational number π, which is pure balderdash. Next, observe that nπ → 0; for, given ǫ > 0, the archimedean property of the real numbers assures us there is a natural number N such that 1 < πǫ so that πn < ǫ for every index n ≥ N. Since { nπ } is a sequence of irrational N numbers converging to the rational number 0, the set of irrational numbers fails to be closed. 2.3.11 For a pair of positive numbers a and b, define sequences {an } and {bn } recursively as follows: define a1 = a and b1 = b. If n is an index for which {an } and {bn } have been defined, define an+1 = an + bn 2 and bn+1 = p an bn . a. Use Exercise 10 to prove that for every index n ≥ 2, an ≥ an+1 ≥ bn+1 ≥ bn . b. Show that the sequences {an } and {bn } converge. Then show that {an } and {bn } have the same limit. Solution √ n ≥ an bn = bn+1 for every index n. Thus a. • By Exercise 10, an+1 = an +b 2 an ≥ bn for every index n ≥ 2. n n = an −b ≥ 0 for every index n ≥ 2, by the above. • an − an+1 = an − an +b 2 2 Thus, an ≥ an+1 for every index n ≥ 2. √ p an bn • bn+1 = = abnn ≥ 1 for every index n ≥ 2 since an ≥ bn for every bn bn index n ≥ 2. Thus, bn+1 ≥ bn for every index n ≥ 2. b. By part (a), {an } is monotonically decreasing and bounded below by b. Thus, there is a number α for which an → α, by the monotone convergence theorem. Similarly, {bn } is monotonically increasing and bounded above by a. Thus, there is a number β for which bn → β, by the monotone convergence theorem. n n → α+β and an+1 = an +b for every index n ≥ 2, it follows that Since an +b 2 2 2 α+β α = 2 whence α = β.