I'm introducing a new series in this post, called "Questions Prompted by Student
Mistakes." I'm deliberately going to write these in more detail that is
necessary so that non-mathematicians can read them.
On a quiz I graded recently, I saw the following:
\[ \cos\left(\frac{2}{x}\right) = \frac{\cos 2}{\cos x} = \frac{2}{x} \]because \( \cos \) totally works like that. So this prompted the
question, is there a function \( f: \mathbb{R} \to \mathbb{R} \)
such that \( \frac{f(a)}{f(b)} = \frac{a}{b} \) for all real
numbers \( a,b \) with \( b \neq 0 \)?
Clearly we can do this for nontrivial linear functions. Every nontrivial linear
function \( f: \mathbb{R} \to \mathbb{R} \) is of the form
\( f(x) = \alpha x \) where \( \alpha \) is your
favorite non-zero constant. We have that
\[ \frac{f(a)}{f(b)} = \frac{\alpha a}{\alpha b} = \frac{a}{b} \]So the question is, then, are linear functions the only ones for which this
property holds? Click the button below to see the answer.
Yes. Suppose \( \frac{f(a)}{f(b)} = \frac{a}{b} \). Then \( f(a) =
\frac{a}{b}\cdot f(b) \), so \( f(a) \) must be a scalar multiple of \(
f(b) \) for each pair of \( a \) and \( b \). In particular, then, for
\( b = 1 \) we have that \( f(a) = \frac{a}{1}\cdot f(1) = af(1)
\). Thus the value of \( f(1) \) determines the entire function. Let \(
f(1) = \alpha \), so then \( f(a) = \alpha a \), and \( f \) is linear.