# Ask Uncle Colin: On the Mediant

Dear Uncle Colin,

Can you explain why $\frac{a+b}{c + d}$ is between $\frac{a}{c}$ and $\frac{b}{d}$?

- Grinding Out Solid Proof Explaining Rationals

Hi, GOSPER, and thanks for your message!

As per usual, there are several methods to show this.

I’m going to assume (since it hasn’t been stated) that $\frac{a}{c}$ is smaller than $\frac{b}{d}$, and that all four of the constants involved are positive.

Also, this $\frac{a+b}{c + d}$ fraction has a name: it’s the mediant.

### Algebraically

From our assumption, we know that $\frac{b}{d} - \frac{a}{c} > 0$, so $cb > ad$.

Consider $\frac{a+b}{c+d} - \frac{a}{c}$. This is equal to $\frac{c(a+b) - a(c+d)}{c(c+d)}$.

Simplifying the top gives $\frac{cb - ad}{c(c+d)}$, which (since $cb > ad$) is positive - so the mediant is greater than $\frac{a}{c}$.

Then consider $\frac{b}{d} - \frac{a+b}{c + d}$. Similarly, this gives $\frac{b(c+d) - d(a+b)}{d(c+d)}$.

Again simplifying the top gives $\frac{bc - ad}{d(c+d)}$, which is also positive - so the mediant is less than $\frac{b}{d}$.

Therefore, the mediant lies between the two fractions.

### Vectors

Consider the points $O (0,0)$, $P(c,a)$ and $Q(d,b)$. The line through the first two has gradient $\frac{a}{c}$; the line through the second has gradient $\frac{b}{d}$.

But if you consider a point $R$ such that $\vec{OR} =\vec{OP} + \vec{OQ}$, it’s clear that $R$ lies above the line through $O$ and $P$, and below the line through $O$ and $Q$. The coordinates of $R$ are $(c+d, a+b)$, so the gradient of the line through $O$ and $R$ is $\frac{a+b}{c+d}$ - and lies between the two other gradients.

Hope that helps!

- Uncle Colin