Mathematics 242 – Principles of Analysis Solutions for Exam 1 – February 22, 2013

I. A) (10) Let

A = ⋃_{n=1}^{∞} (\frac{1}{n}, 2 − \frac{1}{n})

Explain why A is bounded and determine upper and lower bounds for A. Solution: A is bounded below since \frac{1}{n} > 0 for all n. Therefore all elements of A are strictly positive, or x > 0 for all x ∈ A. On the other hand 2 − \frac{1}{n} < 2 for all n, so all elements of A satisfy x < 2.

B) (10) Define: The real number a is a least upper bound of A ⊂ R, and state the Least Upper Bound Axiom for R. Solution: a is a least upper bound of A if (1) a ≥ x for all x ∈ A, and (2) If b ≥ x for all x ∈ A, then b ≥ a. The LUB Axiom states that every nonempty set of real numbers that is bounded above has a least upper bound in R. (The existence is not guaranteed just by defining a term by giving the properties that a lub should satisfy.)

C) (10) Let A be a bounded subset of R and let B = {4 · x − 3 | x ∈ A}. What can be said about lub(B)? Prove your assertion. Solution: If a = lub(A), then we claim lub(B) = 4a − 3. To prove this, note that a exists in R by the LUB Axiom. Since a ≥ x for all x ∈ A, we have 4a ≥ 4x and 4a − 3 ≥ 4x − 3 by properties of the order relation. This shows that 4a − 3 is an upper bound for B. Then, if b is any upper bound for B, we have b ≥ 4x−3 for all x ∈ A, so \frac{b+3}{4} ≥ x for all x ∈ A. This implies \frac{b+3}{4} ≥ a by definition of an lub. Hence b ≥ 4a − 3. This shows that 4a − 3 = 4lub(A) − 3 = lub(B).

II. (20) Let x_{n} be the sequence defined by the rules x_{1} = 1 and x_{n+1} = \frac{1}{3}x_{n} + 1 for all n ≥ 1. Show by mathematical induction that

x_{n} = \frac{1 − \frac{1}{3^{n}}}{1 − \frac{1}{3}} for all n ≥ 1.

Solution: The base case is n = 1. We have x_{1} = 1 = \frac{1− \frac{1}{3}}{1− \frac{1}{3}} , so the formula is true in this case. Now assume that

x_{k} = \frac{1 − \frac{1}{3^{k}}}{1 − \frac{1}{3}} .

By definition,

x_{k+1} = \frac{1}{3}x_{k} + 1 (by definition of the sequence)

= \frac{1}{3} · \frac{1 − \frac{1}{3^{k}}}{1 − \frac{1}{3}} + 1 (by the induction hypothesis)

= \frac{\frac{1}{3} − \frac{1}{3^{k+1}}}{1 − \frac{1}{3}} + 1

= \frac{\frac{1}{3} − \frac{1}{3^{k+1}} + 1 − \frac{1}{3}}{1 − \frac{1}{3}} (common denominator)

= \frac{1 − \frac{1}{3^{k+1}}}{1 − \frac{1}{3}} (by algebra)

This shows that x_{n} is given by the formula above for all n ≥ 1.

III. Let x_{n} = \frac{5^{n}}{7^{n}+3} for all natural numbers n ≥ 1. A) (10) Determine lim_{n→∞} x_{n} intuitively.

Solution: We have

\frac{5^{n}}{7^{n} + 3} = \frac{(5/7)^{n}}{1 + 3(1/7)^{n}}

Since 0 < 5/7 < 1 and 0 < 1/7 < 1, (5/7)^{n} → 0 and (1/7)^{n} → 0. We expect the limit should be 0.

B) (20) Use the ε, n_{0} definition of convergence to prove that {x_{n}} converges to the number you identified in part A. Solution: Given ε > 0, let n_{0} be any natural number satisfying n_{0} > \frac{ln(ε)}{ln(5/7)}. (Note: If ε < 1, then both the top and the bottom of the quotient are negative so the lower bound on n_{0} is positive, and the smaller ε is the larger n_{0} will be.) The for all n ≥ n_{0} we will have

|\frac{5^{n}}{7^{n} + 3} − 0| = \frac{5^{n}}{7^{n} + 3}

< (\frac{5}{7})^{n}

< (\frac{5}{7})^{n_{0}} (since 5/7 < 1 and n ≥ n_{0})

< (\frac{5}{7})^{\frac{ln(ε)}{ln(5/7)}}

= e^{ln(5/7)·ln(ε)/ ln(5/7)}

= e^{ln(ε)} = ε.

This shows that x_{n} → 0.

IV. True-False. For each true statement, give a short proof or reason. For each false statement give an explicit counterexample. A) (10) Let A be a bounded set of real numbers and let a = lub(A). Then for each ε > 0,

there exists x ∈ A such that a − ε < x < a.

Solution: This is FALSE. For example let A = [0, 1] ∪ {2}. Then a = lub(A) = 2. But if ε < 1, then there are no elements of A in the open interval (2 − ε, 2). (The statement would be true if the inequalities were a − ε < x ≤ a, but it is false if we do not allow x = a. )

B) (10) If r is a nonzero rational number and s is a nonzero irrational number, then r/s is irrational. Solution: This is TRUE. Since r ≠ 0 and s ≠ 0, r/s = t ≠ 0. Suppose t is rational. Then we can solve to get s = r/t ∈ Q, which contradicts the assumption s ∉ Q. Hence t must be irrational.

Extra Credit (10) Is it possible to produce a sequence x_{n} whose terms include all the rational numbers p/q with p, q ∈ Z and p, q > 0? If so, give an indication how to construct such a sequence. If not, give a reason why there cannot exist such a sequence.

Solution: There is such a sequence, and we can construct one as follows. List all the ratios of positive integers in a two-dimensional table with denominators constant across the rows and numerators constant along the columns:

\begin{matrix}1/1 & 2/1 & 3/1 & 4/1 & ⋯ \\ 1/2 & 2/2 & 3/2 & 4/2 & ⋯ \\ 1/3 & 2/3 & 3/3 & 4/3 & ⋯ \\ 1/4 & 2/4 & 3/4 & 4/4 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}

We can then construct a sequence by starting at the upper left and listing the elements in this “zig-zag order” (down the first diagonal, up the second, and and alternating that way forever):

x_{1} = 1/1, x_{2} = 2/1, x_{3} = 1/2, x_{4} = 1/3, x_{5} = 2/2, x_{6} = 3/1, x_{7} = 4/1, x_{8} = 3/2, …

Every positive rational will appear somewhere in this sequence (in fact infinitely many times each, do you see why?)