Mathematics 242 – Principles of Analysis Solutions for Exam 1 – February 25, 2011

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

Solution: Every nonempty set of real numbers that is bounded above has a least upper bound. (This is an axiom – we take it as an unproved assertion describing properties of the real number system.)

B) Prove that if A and B are bounded sets of real numbers with lub(A) < lub(B), then there exists a single y ∈ B satisfying y > x for all x ∈ A.

Solution: Since lub(A) < lub(B), lub(A) is not an upper bound for B. Hence there exists at least one y_{0} ∈ B such that lub(A) < y_{0} ≤ lub(B). (The subscript is one way mathematicians often indicate that they are referring to one particular element of a set.) But then, by definition, every x ∈ A satisfies x ≤ lub(A). So x < y_{0} for all x ∈ A.

C) Let A = ∩_{n=1}^{∞} (\frac{−1}{2n}, 1 + \frac{1}{2n}). Explain why A is bounded below and determine glb(A).

Solution: It is certainly true that \frac{−1}{2} is a lower bound for A. But in fact, given any ε > 0, there exist n ∈ N such that \frac{1}{2n} < ε, and hence \frac{−1}{2n} > −ε. We can see that A = [0, 1], so glb(A) = 0.

II. A) Prove that every interval (a, b) with 0 < a < b contains a rational number \frac{m}{n}.

Solution: Since N is not bounded, there exists n ∈ N such that n > \frac{1}{b−a}, or equivalently \frac{1}{n} < b − a. Pick any one such n and fix it. For that fixed n, let M = {m ∈ N | m > na}. This set is not empty (again since N is not bounded). So by the Well Ordering Property, it contains a smallest element. Call that m_{0}. Now m_{0} > na, so \frac{m_{0}}{n} > a. On the other hand, m_{0} − 1 ≠ M, so m_{0} − 1 ≤ na. This implies \frac{m_{0}}{n} ≤ a + \frac{1}{n} < a + b − a = b. Combining these last facts shows that the rational number \frac{m_{0}}{n} ∈ (a, b).

B) Use part A to show that every interval (a, b) with a < b < 0 contains a rational number \frac{m}{n}.

Solution: If a < b < 0, then 0 < (−b) < (−a). By part A, there is a rational number \frac{−m}{n} satisfying (−b) < \frac{−m}{n} < (−a). But then, multiplying these inequalities by −1, a < \frac{m}{n} < b. Since \frac{m}{n} is also rational, we are done.

III. Let f : N → R be defined by f(1) = 2 and f(n + 1) = \frac{2f(n)−1}{3} for all n ≥ 1. Using mathematical induction, show that f(n) > −1 for all n ≥ 1.

Solution: The base case is n = 1. By definition, f(1) = 2 > −1. Now assume as the induction hypothesis that f(k) > −1, and consider f(k + 1) = \frac{2f(k)−1}{3}. Then the properties of order imply

2f(k) > −2
hence 2f(k) − 1 > −3
hence f(k + 1) = \frac{2f(k) − 1}{3} > −1,

which is what we wanted to show. By the principle of mathematical induction, f(n) > −1 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) If A and B are sets of real numbers with the property that a > b for all a ∈ A and all b ∈ B, then glb(A) > lub(B).

Solution: This is FALSE. A counterexample – let B = (−∞, 0) and A = (0, ∞) in R. Then a > b for all a ∈ A and all b ∈ B. But glb(A) = 0 = lub(B).

B) (10) For all n ≥ 1, ∑_{k=0}^{n} \binom{n}{k} \frac{1}{2^{k}} = \frac{3^{n}}{2^{n}}.

Solution: This is TRUE. By the binomial theorem,

∑_{k=0}^{n} \binom{n}{k}x^{k}y^{n−k} = (x + y)^{n}

for all real numbers x, y. We get the desired equality by making x = \frac{1}{2}, y = 1.

C) (10) The smallest c such that {x ∈ R | |x + 1| + |x + 10| = c} is not empty is c = 11.

Solution: This is FALSE. |x + 1| + |x + 10| represents the sum of the distance from x to −1 and the distance from x to −10 along the number line. For any x with −10 ≤ x ≤ −1, the sum of these distances will be 9 < 11.

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 → (−1, 1]? If so, describe one by giving a formula or a graph. If there is no such function, prove it.

Solution: There is such a function. We can construct one as follows: First take any strictly increasing onto function f : R → (−1, 1) such as f(x) = \frac{2}{π} arctan(x). This “misses” 1, of course, so we have to modify f to get 1 in the range. If we define a new function g with g(1) = 1, but g(x) = f(x) for all x ≠ 1, then we have missed f(1) = \frac{1}{2} in the range of f.

But we can make g(2) = f(1) = \frac{1}{2}, then g(3) = f(2), etc. to get a function that is still 1-1 and onto:

\{ \begin{matrix}g(x) = f(x) = \frac{2}{π} arctan(x) & if x ∉ N \\ g(1) = 1 \\ g(n) = f(n − 1) & if n ≥ 2 ∈ N. \end{matrix} .

(This is a mathematical version of the “Hilbert Hotel” with rooms numbered by n ∈ N – there’s always room for one more guest if we shift everyone else down one room!)