I had never heard of the Farey mean, but here it is, brought to you by @howie_hua.
When you add two fractions, you of course remember that you should never just add the numerators and add the denominators across. The resulting fraction will not be the sum of the two addends. But if you do add across (under certain conditions which I’ll show below), the result will be a fraction between the two “addend” fractions. So, you can use the add-across method to find a fraction between two other fractions.
For example, \(\mathtt{\frac{1}{2}+\frac{4}{3}\rightarrow\frac{5}{5}}\). The first “addend” is definitely less than 1, and the second definitely greater than 1. The Farey mean here is exactly 1 (or \(\mathtt{\frac{5}{5}}\)), which is between the two “addend” fractions.
Why Does It Work?
Since this site is fast becoming all linear algebra all the time, let’s throw some linear algebra at this. What we want to show is that, given \(\mathtt{\frac{a}{b}<\frac{c}{d}}\) (we'll go with this assumption for now), \[\mathtt{\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}}\]
for certain positive integer values of \(\mathtt{a,b,c,}\) and \(\mathtt{d}\). I would probably do better to make those inequality signs less-than-or-equal-tos, but let’s stick with this for the present. We’ll start by representing the fraction \(\mathtt{\frac{a}{b}}\) as the vector \(\scriptsize\begin{bmatrix}\mathtt{b}\\\mathtt{a}\end{bmatrix}\) along with the fraction \(\mathtt{\frac{c}{d}}\) as the vector \(\scriptsize\begin{bmatrix}\mathtt{d}\\\mathtt{c}\end{bmatrix}\).
We’re looking specifically at the slopes or angles here (which is why we can represent a fraction as a vector in the first place), so we’ve made \(\scriptsize\begin{bmatrix}\mathtt{d}\\\mathtt{c}\end{bmatrix}\) have a greater slope to keep in line with our assumption above that \(\mathtt{\frac{a}{b}<\frac{c}{d}}\).
The fraction \(\mathtt{\frac{a+c}{b+d}}\) is the same as the vector \(\scriptsize\begin{bmatrix}\mathtt{b+d}\\\mathtt{a+c}\end{bmatrix}\). And since this vector is the diagonal of the vector parallelogram, it will of course have a greater slope than \(\mathtt{\frac{a}{b}}\) but less than \(\mathtt{\frac{c}{d}}\). You can keep going forever—just take one of the side vectors and use the diagonal vector as the other side. So long as you’re making parallelograms, you’ll get a new diagonal shallower than the two side vectors, and the result will be a fraction between the other two.
Incidentally, our assumption at the beginning that \(\mathtt{\frac{a}{b}<\frac{c}{d}}\) doesn't really matter to this picture. If we make \(\mathtt{\frac{c}{d}}\) less than \(\mathtt{\frac{a}{b}}\), our picture simply flips. The diagonal vector still has to be located between the two side vectors.
What Doesn’t Work?
The linear algebra picture of this concept also tells us where this method fails to find a fraction between the two addend fractions. When the two “addend” fractions are equivalent, \(\mathtt{c}\) and \(\mathtt{d}\) are multiples of \(\mathtt{a}\) and \(\mathtt{b}\), respectively, or vice versa. In that case, the resulting fraction looks like this.
The slopes or angles for both addends and for the result are the same, producing a Farey mean that is equal to both fractions.
