Quickly find a number between two others

Here’s a fun observation.

How quickly can you find a number between \(27/97\) and \(27/98\)?

Without using a calculator?

For any two rational numbers¹ you can always find another rational number that lies between them by dividing the sum of their numerators by the sum of their denominators.

Huh? For numbers \(\frac{a}{b}\) and \(\frac{c}{d}\) then \((a + c) / (b + d)\) always lies between them.

In our case: \((27 + 27) / (97 + 98) = 54 / 195\)

(format nil "27 /  97 = ~a~&54 / 195 = ~a  <--~&27 /  98 = ~a~&"
        (/ 27.0 97)
        (/ 54.0 195)
        (/ 27.0 98))

Why do I care?

I was talking to my team at one point and we were making a list of priorities.

We came up with the first couple: 1, 2, 3, … Then came the questions.

What about X? Well, that sits between 1 and 2. Let’s call it 1.5.

What about Y? That sits between 1 and 1.5, so let’s call it 4/3 (1/1 + 3/2 = 4/3) 🤨

What about Z? That sits between 4/3 and 1.5 so let’s call it 7/5 (4/3 + 3/2 = 7/5) 🤨🤨

It then became a bit of a game, testing out my mental arithmetic. They won, I made a mistake, but the underlying algorithm kept working.

Proof(-ish)

Define \(a, b, c, d \in \mathbb{Z}\) and assume without loss of generality

\begin{align} \frac{a}{b} &\leq \frac{c}{d} \nonumber\\ \nonumber\\ \label{eq:order} ad &\leq bc \end{align}

(I’ll leave it to you to verify the following still holds when \(b\) and/or \(d\) are negative.)

If we add \(cd\) to both sides of Eq. \ref{eq:order} we get

\begin{align} ad + cd &\leq bc + cd \nonumber\\ d(a + c) &\leq c(b + d) \nonumber\\ \nonumber\\ \label{eq:cd} \frac{a + c}{b + d} &\leq \frac{c}{d} \end{align}

Equivalently, if we add \(ab\) to both sides of Eq. \ref{eq:order} we find

\begin{align} ad + ab &\leq bc + ab \nonumber\\ a(b + d) &\leq b(a + c) \nonumber\\ \nonumber\\ \label{eq:ab} \frac{a}{b} &\leq \frac{a + c}{b + d} \end{align}

Using Eqs. \ref{eq:cd} and \ref{eq:ab} we get to our desired result

\begin{equation} \frac{a}{b} \leq \frac{a + c}{b + d} \leq \frac{c}{d} \end{equation}