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

$$\begin{array}{rl}\frac{a}{b}& \le \frac{c}{d}\\ \\ \text{(1)}& ad& \le bc\end{array}$$

(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. $\text{1}$ we get

$$\begin{array}{rl}ad+cd& \le bc+cd\\ d(a+c)& \le c(b+d)\\ \\ \text{(2)}& \frac{a+c}{b+d}& \le \frac{c}{d}\end{array}$$

Equivalently, if we add $ab$ to both sides of Eq. $\text{1}$ we find

$$\begin{array}{rl}ad+ab& \le bc+ab\\ a(b+d)& \le b(a+c)\\ \\ \text{(3)}& \frac{a}{b}& \le \frac{a+c}{b+d}\end{array}$$

Using Eqs. $\text{2}$ and $\text{3}$ we get to our desired result

$$\begin{array}{}\text{(4)}& \frac{a}{b}\le \frac{a+c}{b+d}\le \frac{c}{d}\end{array}$$