# 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

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