Mathematics 242 – Principles of Analysis Selected Solutions Midterm Exam 1 – October 4, 2004

II. Consider the following definition: A function f is periodic if and only if there exists a real number P > 0 such that for all real numbers x, f(x + P) = f(x).

B) (10) Give the negation of the statement “f is periodic”.

Solution: The negation of “there exists a real number P > 0 such that for all real numbers x, f(x + P) = f(x)” is: “for all real numbers P > 0, there exists a real number x such that f(x + P) ≠ f(x).” Note: You do not negate the side condition P > 0 when you apply the general rule

∼ (∃P ∋ q(P)) ⇔ ∀P, ∼ q(P)

III. Let A = {x ∈ R : 1 < x^{2} < 9}, let B be the interval [−2, 4), and let C = {1, 2, 3}. A) (10) What is the set A ∩ B? B) (10) What is the set (A ∪ B) \ C? C) (10) Let f : R → R be defined by f(x) = −x + 3. What is the set f^{−1}(A)?

Solution: Since all three of these parts deal with the set A, let’s start by determining exactly which real numbers are in that set! The inequalities 1 < x^{2} < 9 are satisfied if and only if either 1 < x < 3, or −3 < x < −1. Hence A is the union of two open intervals: A = (−3, −1) ∪ (1, 3). Then for part A of the problem,

A ∩ B = {x ∈ R : x ∈ A and x ∈ B} = [−2, −1) ∪ (1, 3)

(that is the set of x with either −2 ≤ x < 1, or 1 < x < 3). For part B of the problem, A ∪ B = (−3, 4), so

(A ∪ B) \ C = {x ∈ R : −3 < x < 4, x ≠ 1, 2, 3} = (−3, 1) ∪ (1, 2) ∪ (2, 3) ∪ (3, 4)

For part C), the function f(x) is defined by f(x) = −x + 3. We want f^{−1}(A) = {x ∈ R : f(x) ∈ A}. This means either −3 < −x + 3 < 1 or 1 < −x + 3 < 3 from the above identification of A as the union (−3, −1) ∪ (1, 3). The first pair of inequalities say −6 < −x < −4 ⇒ 4 < x < 6. The second pair says −2 < −x < 0, so 0 < x < 2. Hence f^{−1}(A) = (0, 2) ∪ (4, 6).

D) (10) Is the function f from part C injective? Is it surjective (onto R)? Justify your answers.

Solution: Even though this part was listed under question III, note that the question does not make reference to the sets A, B, C – they are not relevant in any way here. You must read questions carefully before trying to answer them!! The function f is f(x) = −x + 3. We want to think of this as a function f : R → R. It is injective because

\begin{aligned}f(x_{1}) & = f(x_{2}) \\ ⇒ −x_{1} + 3 & = −x_{2} + 3 \\ ⇒ −x_{1} & = −x_{2} \\ ⇒ x_{1} & = x_{2} \end{aligned}

The function f is also surjective, because given any y ∈ R, we have

y = f(x) = −x + 3 ⇔ x = −(y − 3) = −y + 3

So y = f(x) for x = −y + 3. (In other words, f is actually its own inverse function(!))

IV. (20) Show that the open interval (0, 1) in the real numbers is not a denumerable set.

See the class notes and Theorem 8.12 in the book for this. I was seeing a lot of memorization without much understanding on this question(!)

V. (15) Prove that for all n ∈ N,

1^{3} + 2^{3} + ⋯ + n^{3} = \frac{1}{4}n^{2}(n + 1)^{2}.

Solution: We use mathematical induction. The base case n = 1 is true because 1^{3} = 1 = \frac{1}{4}1^{2}(1 + 1)^{2}. For the induction step, assume the result is true for n = k, and consider the sum of the cubes of the first k + 1 integers. By the induction hypothesis, we have

1^{3} + 2^{3} + ⋯ + k^{3} + (k + 1)^{3} = \frac{1}{4}k^{2}(k + 1)^{2} + (k + 1)^{3}

= (k + 1)^{2} (\frac{1}{4}k^{2} + k + 1)

= \frac{1}{4}(k + 1)^{2}(k^{2} + 4k + 4)

= \frac{1}{4}(k + 1)^{2}(k + 2)^{2}

= \frac{1}{4}(k + 1)^{2}((k + 1) + 1)^{2}

which is what we wanted to show. Hence the formula for the sum of the cubes works for all n ∈ N.

Extra Credit

(10) Let x be a non-negative real number. Show that if x ≤ ε for all ε > 0, then x = 0.

Solution: We will prove the contrapositive form: If x ≠ 0, then there exists a ε > 0 such that x > ε. We are still working under the hypothesis that x is non-negative. So if x ≠ 0, then by the trichotomy property of the order relation (O1 in the book), we have x > 0 (strictly positive). But then there certainly is a positive real number ε such that 0 < ε < x (take ε = x/2 for instance).