How I Impressed My Wife: Part 2c

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

In this series, I’ll explore different ways of evaluating this integral.So far in this series, I’ve shown that

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

I now employ the substitution $u = \tan x$ , so that $du = \sec^2 x dx$ . Also, the endpoints change from $-\pi/2 < x < \pi/2$ to $-\infty < u < \infty$ , so that

$Q = 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$

I’ll continue with the evaluation of this integral in tomorrow’s post.

How I Impressed My Wife: Part 2b

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

In this series, I’ll explore different ways of evaluating this integral.In yesterday’s post, I showed that

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

We now multiply the top and bottom of the integrand by $\sec^2 x$ . This is permissible because $\sec^2 x$ is defined on the interior of the interval $(-\pi/2, \pi/2)$ — which is why I needed to adjust the limits of integration in the first place. I obtain

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{\cos^2 x \sec^2 x + 2 a \sin x \cos x \sec^2 x + (a^2 + b^2) \sin^2 x \sec^2 x}$

Next, I use some trigonometric identities to simplify the denominator:

$\cos^2 x \sec^2 x = \cos^2 x \displaystyle \frac{1}{\cos^2 x} = 1$
$\sin x \cos x \sec^2 x = \sin x \cos x \frac{1}{\cos^2 x} = \displaystyle \frac{\sin x}{\cos x} = \tan x$
$\sin^2 x \sec^2 x = \sin^2 x \displaystyle \frac{1}{\cos^2 x} = \displaystyle \left( \frac{\sin x}{\cos x} \right)^2 = \tan^2 x$

Therefore, the integral becomes

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

I’ll continue with the evaluation of this integral in tomorrow’s post.

How I Impressed My Wife: Part 2a

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

In this series, I’ll explore different ways of evaluating this integral.

I begin by adjusting the range of integration:

$Q = Q_1 + Q_2 + Q_3$ ,

where

$Q_1 = \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ ,

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ ,

$Q_3 = \displaystyle \int_{3\pi/2}^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ .

I’ll begin with $Q_3$ and apply the substitution $u = x - 2\pi$ , or $x = u + 2\pi$ . Then $du = dx$ , and the endpoints change from $3\pi/2 \le x 2\pi$ to $-\pi/2 \le u \le 0$ . Therefore,

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 (u+2\pi) + 2 a \sin (u+2\pi) \cos (u+2\pi) + (a^2 + b^2) \sin^2 (u+2\pi)}$ .

Next, we use the periodic property for both sine and cosine — $\sin(x + 2\pi) = \sin x$ and $\cos(x + 2\pi) = \cos x$ — to rewrite $Q_3$ as

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}$ .

Changing the dummy variable from $u$ back to $x$ , we have

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ .

Therefore, we can combined $Q_3 + Q_1$ into a single integral:

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$+ \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Next, we work on the middle integral $Q_2$ . We use the substitution $u = x - \pi$ , or $x = u + \pi$ , so that $du = dx$ . Then the interval of integration changes from $\pi/2 \le x \le 3\pi/2$ to $-\pi/2 \le u \le \pi/2$ , so that

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 (u+\pi) + 2 a \sin (u+\pi) \cos (u+\pi) + (a^2 + b^2) \sin^2 (u+\pi)}$ .

Next, we use the trigonometric identities

$\sin(u + \pi) = \sin u \cos \pi + \cos u \sin \pi = \sin u \cdot (-1) + \cos u \cdot 0 = - \sin u$ ,

$\cos(u + \pi) = \cos u \cos \pi - \sin u \sin \pi = \cos u \cdot (-1) - \sin u \cdot 0 = - \cos u$ ,

so that the last integral becomes

$Q_2 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{(-\cos u)^2 + 2 a (-\sin u)(- \cos u) + (a^2 + b^2) (-\sin u)^2}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

On the line above, I again replaced the dummy variable of integration from $u$ to $x$ . We see that $Q_2 = Q_1 + Q_3$ , and so

$Q = Q_1 + Q_2 + Q_3$

$Q = 2 Q_2$

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

I’ll continue with the evaluation of this integral in tomorrow’s post.

Functions that commute (Part 2)

Math With Bad Drawings had a nice post with pedagogical thoughts on the tendency of students to commute two functions that don’t commute:

The author’s proposed remedies:

Teach the distributive law more carefully. Draw pictures. Work examples. Talk about “bags.” Make sure they understand the meaning behind this symbolism.

Teach function notation much more carefully. Give them the chance to practice it. Think like Dan Meyer and seek activities that create the intellectual need for function notation.

Keep stamping out the “everything is linear” error when it crops up. Like the common cold, it’ll probably never be entirely eradicated, but good mathematical hygiene should reduce its prevalence.

I agree with all three points. Concerning the third point, here’s an earlier post of mine concerning these kinds of mistakes (and others), with a one-liner I’ll use to try to get students to remember not to make these kinds of mistakes:

I wish I could remember the speaker’s name, but I heard the following one-liner at a state mathematics conference many years ago, and I’ve used it to great effect in my classes ever since. Whenever I present a property where two functions commute, I’ll say, “In other words, the order of operations does not matter. This is a big deal, because, in real life, the order of operations usually is important. For example, this morning, you probably got dressed and then went outside. The order was important.”
If that fails, then I’ll cite Finding Nemo, trying to minimize frustration by keeping the mood light.

And if that fails, I’ll cite The Princess Bride. One of the most common student mistakes with logarithms is thinking that
$\log_b(x+y) = \log_b x + \log_b y$ .

When I first started my career, I referred to this as the Third Classic Blunder. The first classic blunder, of course, is getting into a major land war in Asia. The second classic blunder is getting into a battle of wits with a Sicilian when death is on the line. And the third classic blunder is thinking that $\log_b(x+y)$ somehow simplfies as $\log_b x + \log_b y$ .

Sadly, as the years pass, fewer and fewer students immediately get the cultural reference. On the bright side, it’s also an opportunity to introduce a new generation to one of the great cinematic masterpieces of all time.

Is 2i less than 3i? (Part 4: Two other attempted inequalities)

In yesterday’s post, I demonstrated that there is no subset $\mathbb{C}^+ \subset \mathbb{C}$ of the complex numbers which satisfies the following four axioms:

If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1+z_2 \in \mathbb{C}^+$
If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1 z_2 \in \mathbb{C}^+$ .
For every $z \ne 0$ , either $z \in \mathbb{C}^+$ or $-z \in \mathbb{C}^+$ , but not both.
$0 \notin \mathbb{C}^+$

However, it’s instructive (and fun) to try to construct such a set. Yesterday I showed that the following subset satisfies three of the four axioms:

$\mathbb{C}^+ = \{x + iy : x > 0 \qquad \hbox{or} \qquad x = 0, y > 0\}$

Apostol’s calculus suggests two other subsets to try:

$\mathbb{C}^+ = \{x + iy : x^2 + y^2 > 0 \}$

and

$\mathbb{C}^+ = \{x + iy : x > y\}$

Neither of these sets work either, but I won’t spoil the fun for you by giving you the proofs. I leave a thought bubble if you’d like to try to figure out which of the four axioms are satisfied by these two notions of “positive” complex numbers.

Is 2i less than 3i? (Part 3: An inequality that almost works)

In yesterday’s post, I demonstrated that there is no subset $\mathbb{C}^+ \subset \mathbb{C}$ of the complex numbers which satisfies the following four axioms:

If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1+z_2 \in \mathbb{C}^+$
If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1 z_2 \in \mathbb{C}^+$ .
For every $z \ne 0$ , either $z \in \mathbb{C}^+$ or $-z \in \mathbb{C}^+$ , but not both.
$0 \notin \mathbb{C}^+$

However, it’s instructive (and fun) to try to construct such a set. One way of attempting this is defining

$\mathbb{C}^+ = \{x + iy : x > 0 \qquad \hbox{or} \qquad x = 0, y > 0\}$

This set $\mathbb{C}^+$ leads to the lexicographic ordering of the complex numbers: if $z_1 = a_1 + i b_1$ and $z_2 = a_2 + i b_2$ , where $a_1, a_2, b_1, b_2 \in \mathbb{R}$ , we say that $z_1 \prec z_2$ if

$a_1 < a_2 \qquad \hbox{or} \qquad a_1 = a_2, b_1 < b_2$

I used the symbol $z_1 \prec z_2$ because, as we’ll see, $\prec$ satisfies some but not all of the usual properties of an inequality. This ordering is sometimes called the “dictionary” order because the numbers are ordered like the words in a dictionary… the real parts are compared first, and then (if that’s a tie) the imaginary parts are compared. See Wikipedia and Mathworld for more information.

In any case, defining $\mathbb{C}^+$ in this way satisfies three of the four order axioms.

Suppose . It’s straightforward to show that . Let and , where . Then , and so .
- Case 1: If $a_1 + a_2 > 0$ , then clearly $z_1 + z_2 \in \mathbb{C}^+$ .
- Case 2: If $a_1 + a_2 = 0$ , that’s only possible if $a_1 = 0$ and $a_2 = 0$ . But since $z_1, z_2 \in \mathbb{C}^+$ , that means that $b_1 > 0$ and $b_2 > 0$ . Therefore, $b_1 + b_2 > 0$ . Since $a_1 + a_2 = 0$ , we again conclude that $z_1 + z_2 \in \mathbb{C}^+$ .
Suppose , where . Then or . We now show that, no matter what, or , but not both.
- Case 1: If $a > 0$ , then $-a < 0$ , and so $z \in \mathbb{C}^+$ but $-z \notin \mathbb{C}^+$ .
- Case 2: If $a < 0$ , then $-a > 0$ , and so $-z \in \mathbb{C}^+$ but $z \notin \mathbb{C}^+$ .
- Case 3: If , then since . Also, if , then , so that and .
  - Subcase 3A: If $b > 0$ , then $-b < 0$ , and so $z \in \mathbb{C}^+$ but $-z \notin \mathbb{C}^+$ .
  - Subcase 3B: If $b < 0$ , then $-b > 0$ , and so $-z \in \mathbb{C}^+$ but $z \notin \mathbb{C}^+$ .
By definition, $0 = 0 + 0i \notin \mathbb{C}^+$ .

However, the fourth property fails. By definition, $i = 0 + 1i \in \mathbb{C}^+$ . However, $i \cdot i = -1 + 0i \notin \mathbb{C}^+$ .

Because this definition of $\mathbb{C}^+$ satisfies three of the four order axioms, the relation $\prec$ satisfies some but not all of the theorems stated in the first post of this series. For example, if $z_1 \prec z_2$ and $z_2 \prec z_3$ , then $z_1 \prec z_3$ . Also, if $z_1 \prec z_2$ and $w_1 \prec w_2$ , then $z_1 + w_1 \prec z_2 + w_2$ .

I’ll leave it to the interested reader to determine which of the theorems are true, and which are false (and have counterexamples).

Is 2i less than 3i? (Part 2: Proof by contradiction)

Is $2i$ less than $3i$ ?

That’s a very natural question for a student to ask when first learning about complex numbers. The short answer is, “No… there isn’t a way to define inequality for complex numbers that satisfies all the properties of real numbers.”

I demonstrated this in a roundabout way in yesterday’s post. Today, let’s tackle the issue using a proof by contradiction. (I almost said, “Let’s tackle the issue more directly,” but avoided that phrase because a proof by contradiction is sometimes called an indirect proof, which would lead to the awkward sentence “Let’s tackle the issue more directly with an indirect proof.”)

Suppose that there’s a subset $\mathbb{C}^+ \subset \mathbb{C}$ of the complex numbers which satisfies the following four axioms:

If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1+z_2 \in \mathbb{C}^+$
If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1 z_2 \in \mathbb{C}^+$ .
For every $z \ne 0$ , either $z \in \mathbb{C}^+$ or $-z \in \mathbb{C}^+$ , but not both.
$0 \notin \mathbb{C}^+$

As discussed yesterday, the ordinary properties of inequalities derive from these four order axioms. Assuming these four order axioms are true for the complex numbers, let’s investigate whether $i$ is “positive” or not. According to the third axiom, either $i \in \mathbb{C}^+$ or $-i \in \mathbb{C}^+$ , but not both.

Case 1. $i \in \mathbb{C}^+$ . Then by the second axiom, $i \cdot i \in \mathbb{C}^+$ , or $-1 \in \mathbb{C}^+$ . Applying the second axiom again, since $i \in \mathbb{C}^+$ and $-1 \in \mathbb{C}^+$ , we have $i \cdot (-1) \in \mathbb{C}^+$ , or $-i \in \mathbb{C}^+$ . But that’s impossible because we assumed that $i \in \mathbb{C}^+$ .

Case 2. $-i \in \mathbb{C}^+$ . Then by the second axiom, $(-i) \cdot (-i) \in \mathbb{C}^+$ , or $-1 \in \mathbb{C}^+$ . Applying the second axiom again, since $-i \in \mathbb{C}^+$ and $-1 \in \mathbb{C}^+$ , we have $(-i) \cdot (-1) \in \mathbb{C}^+$ , or $i \in \mathbb{C}^+$ . But that’s impossible because we assumed that $-i \in \mathbb{C}^+$ .

Either way, we obtain a contradiction. Therefore, there is no subset $\mathbb{C}^+$ of the complex numbers that serves as the “positive” complex numbers, and so there’s no way to define inequalities for complex numbers that satisfies all of the usual properties of inequalities.

Is 2i less than 3i? (Part 1: Order Axioms)

Is $2i$ less than $3i$ ?

However, for this answer to make sense, we need to talk about how inequalities are defined in the first place.

Following Apostol’s calculus, there are four axioms from which the ordinary notions of inequality follow. We shall assume that there exists a certain subset $\mathbb{R}^+ \subset \mathbb{R}$ , called the set of positive numbers, which satisfies the following four order axioms:

If $x, y \in \mathbb{R}^+$ , then $x+y \in \mathbb{R}^+$
If $x, y \in \mathbb{R}^+$ , then $xy \in \mathbb{R}^+$ .
For every real $x \ne 0$ , either $x \in \mathbb{R}^+$ or $-x \in \mathbb{R}^+$ , but not both.
$0 \notin \mathbb{R}^+$

We then define the symbols $<$ , $\le$ , $>$ , and $\ge$ in the obvious way:

$x < y$ means that $y-x$ is positive.
$x > y$ means that $y < x$.
$x \le y$ means that either $x < y$ or $x=y$ .
$x \ge y$ means that $y \le x$

From only these four axioms, many familiar theorems about inequalities can be proven. For what it’s worth, when I was a student in Algebra I, I had to prove nearly all of these theorems.

If $a, b \in \mathbb{R}$ , then exactly one of the following three relations is true: $a < b$ , $a > b$ , $a = b$ .
If $a < b$ and $b < c$ , then $a < c$ .
If $a < b$ , then $a + c < b + c$ .
If $a < b$ and $c > 0$ , then $ac < bc$ .
If $a \ne 0$ , then $a^2 > 0$ .
$1 > 0$ .

We interrupt this list with a public-service announcement: Yes, there’s a proof that $1$ is greater than $0$ . When I tell this to students, I can usually see their heads start to spin, as they think, “Of course we know that!” Then I ask them what the definitions of $1$ and $0$ are. Usually, they have no idea. Then I’ll remind them that $0$ is defined to be the additive identity (so that $x + 0 = x$ and $0 + x = x$ for all real numbers $x$ ), while $1$ is defined to be the multiplicative identity (so that $x \cdot 1 = x$ and $1 \cdot x = x$ for all real numbers $x$ ). Based on those definitions alone, I then ask my students, is it obvious that the multiplicative identity has to be larger than the additive identity? The answer is no, which is why the above order axioms are needed.

Here are some more familiar theorems about inequalities that derive from the four order axioms.

If $a < b$ and $c < 0$ , then $ac > bc$ .
If $a < b$ , then $-a > -b$ .
If $a < 0$ , then $-a > 0$ .
If $ab > 0$ , then $a$ and $b$ are either both positive or both negative.
If $a < c$ and $b < d$ , then $a + b < c + d$ .
If $a < 0$ and $b < 0$ , then $a + b < 0$ .
If $a > 0$ , then $1/a > 0$ .
If $a < 0$ , then $1/a < 0$ .
If $0 < a < b$ , then $0 < 1/b < 1/a$ .
If $a \le b$ and $b \le c$ , then $a \le c$ .
If $a \le b \le c$ and $a = c$ , then $b = c$ .

These last two theorems are less familiar. They basically state that (1) there is no “biggest” real number and (2) there is no positive number that’s immediately to the right of 0.

There is no real number $a$ so that $x \le a$ for all real numbers $x$ .
If $0 \le x < h$ for every positive real number $h$ , then $x =0$ .

Here’s another important theorem that ultimately derives from the four order axioms, proving that a number system including $\sqrt{-1}$ is incompatible with the four order axioms.

There is no real number $x$ so that $x^2 + 1 = 0$ .

The proof of this theorem is simple, given the theorems above. If $x = 0$ , then $x^2 + 1 = 1$ , which is greater than 0. If $x \ne 0$ , then $x^2 > 0$ , and so $x^2 + 1 > 0 + 1 > 1$ . Since $1 > 0$ , it follows by transitivity that $x^2 > 0$ . Either way, $x^2 + 1 > 0$ , and so $x^2 + 1 \ne 0$ .

Proving theorems and special cases (Part 16): An old homework problem

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

The following problem appeared on a homework assignment of mine about 30 years ago when I was taking Honors Calculus out of Apostol’s book. I still remember trying to prove this theorem (at the time, very unsuccessfully) like it was yesterday.

Theorem. If $f(x)$ is a continuous function so that $f(x+y) = f(x) + f(y)$ , then $f(x) = cx$ for some constant $c$ .

Proof. The proof mirrors that of the uniqueness of the logarithm function, slowly proving special cases to eventually prove the theorem for all real numbers $x$ .

Case 1. $x = 0$ . If we set $x =0$ and $y = 0$ , then

$f(0+0) = f(0) + f(0)$

$f(0) = 2 f(0)$

$0 = f(0)$

Case 2. $x \in \mathbb{N}$ . If $x$ is a positive integer, then

$f(x) = f(1 + 1 + \dots + 1)$

$f(x) = f(1) + f(1) + \dots + f(1)$

$f(x) = xf(1)$ .

(Technically, this should be proven by induction, but I’ll skip that for brevity.) If we let $c = f(1)$ , then $f(x) = cx$ .

Case 3. $x \in \mathbb{Z}$ . If $x$ is a negative integer, let $x = -n$ , where $n$ is a positive integer. Then

$f(x + (-x)) = f(x) + f(-x)$

$f(0) = f(x) + f(n)$

$0 = f(x) + cn$

$-cn = f(x)$

$cx = f(x)$

Case 4. $x \in \mathbb{Q}$ . If $x$ is a rational number, then write $x = p/q$ , where $p$ and $q$ are integers and $q$ is a positive integer. We’ll use the fact that $p = xq = p/q \times q = p/q + p/q + \dots + p/q$ , where the sum is repeated $q$ times.

$f(p/q + p/q + \dots + p/q) = f(p)$

$f(p/q) + f(p/q) + \dots + f(p/q) = cp$

$q f(p/q) = cp$

$f(p/q) = cp/q$

$f(x) = cx$

Case 5. $x \in \mathbb{R}$ . If $x$ is a real number, then let $\{r_n\}$ be a sequence of rational numbers that converges to $x$ , so that

$\lim_{n \to \infty} r_n = x$

Then, since $f$ is continuous,

$f(x) = f \left( \displaystyle \lim_{n \to \infty} r_n \right)$

$f(x) =\displaystyle \lim_{n \to \infty} f(r_n)$

$f(x) = \displaystyle \lim_{n \to \infty} c r_n$

$f(x) = cx$

QED

Random Thought #1: The continuity of the function $f$ was only used in Case 5 of the above proof. I’m nearly certain that there’s a pathological discontinuous function that satisfies $f(x+y) = f(x) + f(y)$ which is not the function $f(x) = cx$ . However, I don’t know what that function might be.

Random Thought #2: For what it’s worth, this same idea can be used to solve the following problem that was posed during UNT’s Problem of the Month competition in January 2015. I won’t solve the problem here so that my readers can have the fun of trying to solve it for themselves.

Problem. Determine all nonnegative continuous functions that satisfy

$f(x+t) = f(x) + f(t) + 2 \sqrt{f(x)} \sqrt{f(t)}$ .

Proving theorems and special cases (Part 13): Uniqueness of logarithms

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student: