Math 312, Intro. to Real Analysis: Midterm Exam #1 Solutions

Stephen G. Simpson

Friday, February 13, 2009

1. True or False (3 points each)

(a) Every ordered field has the Archimedean property.

Answer: False.

(b) The ordered field axioms imply |a − b| ≤ |a| + |b| for all a, b. Answer: True.

(c) If lim a_{n} = −∞ then lim sup a_{n} = −∞.

Answer: True.

(d) For any sequence of real numbers, the lim inf and the lim sup always exist and furthermore the lim inf is always ≤ the lim sup. Answer: True.

(e) The equation 3x^{3} + 2x^{2} + 3x + 2 = 0 has a rational solution. Answer: True. A rational solution is x = −2/3.

(f) ^{3} \sqrt{216} is an irrational number. Answer: False. ^{3} \sqrt{216} = 6.

(g) The limit of a convergent sequence of negative numbers is negative. Answer: False. For example, the sequence −1/n for n = 1, 2, 3, … has 0 as its limit.

(h) The limit of a convergent sequence of rational numbers is rational. Answer: False.

(i) Every interval contains at least three rational numbers. Answer: True.

(j) Every bounded sequence of real numbers is convergent. Answer: False.

(k) Every convergent sequence of real numbers is bounded. Answer: True.

(l) Every monotone sequence of real numbers is convergent. Answer: False. For example, the sequence 1, 2, 3, … is divergent to ∞.

(m) If (a_{n}) is a monotone sequence of real numbers, then lim a_{n} exists and belongs to the interval (−∞, ∞). Answer: False. Same example as above.

2. (7 points each)

(a) Give an example of a sequence of real numbers such that

−∞ < inf a_{n} < lim a_{n} < sup a_{n} < ∞.

Answer: An example is a_{n} = \frac{(−1)^{n}}{n}. The inf is −1, the sup is 1/2, and the limit is 0.

(b) Give an example of a sequence of real numbers such that

lim sup a_{n}, lim inf a_{n}, sup a_{n}, inf a_{n}

are four distinct real numbers.

Answer: An example is a_{n} = (−1)^{n} (1 + \frac{1}{n}). The inf is −2, the sup

is 1 + 1/2, the lim inf and lim sup are −1 and 1.

(c) Give an example of a sequence of real numbers such that

lim inf a_{n} = −∞ and lim sup a_{n} = \sqrt{2}.

Answer: An example is −1, \sqrt{2}, −2, \sqrt{2}, −3, \sqrt{2}, −4, …. In other words, a_{2n} = \sqrt{2} and a_{2n−1} = −n for all n.

3. (8 points) It can be shown that ^{3} \sqrt{1 + \sqrt{5}} is an algebraic number, i.e., it is a solution of some polynomial equation with integer coefficients. Find such an equation.

Answer: Let α = ^{3} \sqrt{1 + \sqrt{5}}. Then α^{3} = 1 + \sqrt{5}, i.e., α^{3} − 1 = \sqrt{5} hence α^{6} − 2α^{3} + 1 = 5, i.e., α^{6} − 2α^{3} − 4 = 0. The desired equation is x^{6} − 2x^{3} + 4 = 0.

4. (8 points) Find all candidates for rational solutions of the equation

2x^{2} − ax + 5 = 0

where a is an unspecified integer.

Answer: The candidates are ±5, ±\frac{5}{2}, ±1, ±\frac{1}{2}. This is according to the Rational Zeros Theorem, page 9 in the Ross textbook.

5. (12 points) Use algebra plus limit laws to calculate

lim \frac{\sqrt{2n^{2} + 5n}}{n + 4}.

Answer:

lim \frac{\sqrt{2n^{2} + 5n}}{n + 4} = lim \frac{\sqrt{2 + \frac{5}{n}}}{1 + \frac{4}{n}} = \frac{lim \sqrt{2 + \frac{5}{n}}}{lim 1 + \frac{4}{n}}

= \frac{\sqrt{lim 2 + \frac{5}{n}}}{lim 1 + \frac{4}{n}} = \frac{\sqrt{2 + lim \frac{5}{n}}}{1 + lim \frac{4}{n}}
= \frac{\sqrt{2 + 0}}{1 + 0}  = \sqrt{2}.

6. (12 points) It can be shown that

lim \frac{^{3} \sqrt{n} − 5001}{^{3} \sqrt{n} − 1001} = 1.

Given ϵ > 0, find an N so large that

| \frac{^{3} \sqrt{n} − 5001}{^{3} \sqrt{n} − 1001} − 1| < ϵ

holds for all n > N. Answer: Given ϵ > 0, we want

| \frac{^{3} \sqrt{n} − 5001}{^{3} \sqrt{n} − 1001}| < ϵ

and by algebra this is equivalent to saying that

| \frac{−4001}{^{3} \sqrt{n} − 1001}| < ϵ.

If n > 1001^{3} then ^{3} \sqrt{n} − 1001 > 0, and | − 4001| = 4001, so the above is equivalent to saying that

\frac{4001}{^{3} \sqrt{n} − 1001} < ϵ.

By algebra, this is equivalent to

\frac{4001}{ϵ} + 1001 < ^{3} \sqrt{n}

i.e.,

n > (\frac{4001}{ϵ} + 1001)^{3} .

Thus we may take

N = (\frac{4001}{ϵ} + 1001)^{3} .