Mathematics 242 – Principles of Analysis Solutions for Exam 1 – October 7, 2005

I. A) (10) State the Axiom of Completeness for the real number system.

Solution: Every set A ⊂ R that is bounded above has a least upper bound s = sup(A) in R.

B) (10) Show that if s = sup(A) for some A ⊂ R and ε > 0, then there exists some a ∈ A satisfying s − ε < a ≤ s.

Solution: By definition, s is the real smallest number that is an upper bound for A. Since ε > 0, s − ε < 0 and hence s − ε is not an upper bound for A. This implies that there is some a ∈ A with s − ε < a. a ≤ s because s is an upper bound for A.

C) (10) Let A = ∪_{n=1}^{∞} (0, 1 − \frac{1}{2n}). Explain why A is bounded above and determine sup(A).

Solution: If x ∈ A, then there is some n such that x < \frac{1}{2n}. Since \frac{1}{2n} < 1 for all n ∈ N, we get x < 1 for all x ∈ A. In fact, sup(A) = 1 since if y = 1 − ε < 1, then the Archimedean Property implies that there exists n ∈ N such that \frac{1}{2n} < ε, so 1 − \frac{1}{2n} > 1 − ε = y. Hence there exist x ∈ A with x > y too, and y cannot be an upper bound for A. Since 1 is an upper bound for A, but no y < 1 is, sup(A) = 1.

II. A) (5) Find

lim_{n→∞} \frac{2n^{2} − 1}{5n^{2} + 2n}

using the Algebraic Limit Theorem.

Solution:

lim_{n→∞} \frac{2n^{2} − 1}{5n^{2} + 2n} = lim_{n→∞} \frac{2 − \frac{1}{n^{2}}}{5 + \frac{2}{n}} = \frac{2}{5}

B) (15) Prove that x_{n} = \frac{2n^{2}−1}{5n^{2}+2n} converges to your limit from part A using the definition.

Solution: Preliminaries: We can use some slightly tricky estimates to reduce the complication of the N that works for a given ε > 0:

|\frac{2n^{2} − 1}{5n^{2} + 2n} − \frac{2}{5}| = \frac{4n + 5}{25n^{2} + 10n}

< \frac{4n + 5}{25n^{2}}

< \frac{5n + 5}{25n^{2}}

= \frac{n + 1}{5n^{2}}

< \frac{2n}{5n^{2}} = \frac{2}{5n}

Proof: Let ε > 0, and let N ∈ N satisfy N > \frac{2}{5ε}. Then for all n ≥ N,

|\frac{2n^{2} − 1}{5n^{2} + 2n} − \frac{2}{5}| = \frac{4n + 5}{25n^{2} + 10n}

< \frac{4n + 5}{25n^{2}}

< \frac{5n + 5}{25n^{2}}

= \frac{n + 1}{5n^{2}}

< \frac{2n}{5n^{2}}

≤ \frac{2}{5N}

< ε

Hence lim_{n→∞} \frac{2n^{2}−1}{5n^{2}+2n} = \frac{2}{5}. III. Let (x_{n}) be the sequence defined by x_{1} = 2 and x_{n+1} = \frac{2x_{n}−1}{3} for all n ≥ 1. A) (15) Using mathematical induction, show that x_{n} ≥ −1 for all n ≥ 1. B) (15) Using mathematical induction, show that (x_{n}) is strictly decreasing: x_{n+1} < x_{n}

for all n ≥ 1. IV. True-False. For each true statement give a short proof or reason; for each false statement give a counterexample. A) (10) If A and B are sets of real numbers with the property that a > b for all a ∈ A

and b ∈ B, then inf(A) > sup(B). B) (10) If r is a nonzero rational number and t is an irrational number, then rt is irrational. Extra Credit (10) Caution: this problem may be “habit forming.” Only attempt after finishing the rest of the exam! Can you find a 1-1, onto function f : R → [0, 1]? If so, describe one by giving a formula or a graph. If not, say why not.