Naively adding fractions

Wayne Junn Voo

13 November 2026

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Suppose we wanted to add $a/b$ and $c/d$. We were always taught to add them by first finding a common denominator, then adding the numerators, so something like $a/b+c/d=(ad+bc)/bd$. If you were feeling rebellious, you probably thought at some point: What is the problem with naively adding them together? Why can we not add them like $a/b + c/d =(a+c)/(b+d)?$ It turns out that we can, in fact, do that, and it leads to a lot of interesting maths. This so-called ‘naively defined addition’ opens up the door to a field in number theory called Diophantine approximation.

A new game

So what is Diophantine approximation all about? Here is the game we are playing. You give me your favourite real number (preferably irrational), say $\alpha$. My goal is to get reasonably close to it, or to approximate it using rational numbers with relatively small denominators. The small denominator restriction here is where the real fun lies; otherwise, I could just define a sequence of rationals that converge to your real number since the rationals are dense in the reals (which really is just a fancier way of saying ‘if you give me any two real numbers then I can give you a rational number in between them’), and that is just no fun.

After giving this question some thought, you could probably come up with a bound like this. First, we mark every rational with denominator $q$ on the number line. We know that $\alpha$ has to lie somewhere on the number line. In particular, it has to lie in between two consecutive fractions with denominator $q$, so the distance between $\alpha$ and any one of these fractions is upper bounded by $1/q$. Now I am either closer to one fraction or the other, so by picking the closer one to be our approximation, we get $1/2q$. So pictorially, we have something like this:

We can write this bound more formally: let $\alpha$ be a real number, then for each natural number $q$ there is some integer $p$ such that $$ \left| \alpha – \frac{p}{q}\right| \leq \frac{1}{2q}. $$ Here’s another, perhaps more intuitive, way to think about it. If we multiply the whole expression by $q$, we get $$ \left| q\alpha – p \right| \leq \frac{1}{2}. $$ In some sense, we can accurately determine what our $p$ needs to be, given $\alpha$ and $q$. It has to either be the floor or the ceiling of $q \alpha$!

Now we ask ourselves, can we do better? In other words, can I make the upper bound smaller, given some $\alpha$ and $q$?

Just naively add the fractions

Let’s think back to our ill-defined addition. Before getting too excited, let’s restrict ourselves to fractions in between 0 and 1 for now. We are going to create a sequence by adding fractions. So let’s start with the sequence
$$ \left( \frac{0}{1},\frac{1}{1} \right). $$
Now let’s add these two fractions using our somewhat diabolical definition of addition. We have
$0/1 + 1/1 = (0+1)/(1+1) = 1/2$. Now inserting $1/2$ in between them, we get $$ \left( \frac{0}{1}, \frac{1}{2},\frac{1}{1} \right). $$
Doing this again, $$ \left( \frac{0}{1},\frac{1}{3}, \frac{1}{2}, \frac{2}{3},\frac{1}{1} \right), $$
and again, we get $$ \left( \frac{0}{1},\frac{1}{4} ,\frac{1}{3}, \frac{2}{5},\frac{1}{2},\frac{3}{5} , \frac{2}{3}, \frac{3}{4},\frac{1}{1} \right). $$

Notice that, in terms of their absolute values, the fractions appear to be automatically arranged in ascending order, and they also appear to be irreducible by default. Before the list gets too long, we let $\mathfrak{F}_N$ denote the sequence generated as above, with the restriction that all elements have denominators less than or equal to $N$. So we define $\mathfrak{F}_N$ to be the ordered sequence of irreducible rational numbers with denominators $\leq N$. For example,

\begin{align*} \mathfrak{F}_3&= \left( \frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{1}{1} \right), \\ \mathfrak{F}_4&= \left( \frac{0}{1},\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3}, \frac{3}{4},\frac{1}{1} \right), \\ \mathfrak{F}_5&= \left( \frac{0}{1},\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3}, \frac{3}{4},\frac{4}{5},\frac{1}{1} \right). \end{align*}

We call $\mathfrak{F}_N$ the Farey sequence of order N, and the fractions Farey fractions. Here’s an interesting observation. If we want to generate $\mathfrak{F}_{N+1}$, we can start from $\mathfrak{F}_N$. We then find two consecutive fractions in $\mathfrak{F}_N$, say $a/b$ and $c/d$. Now we add them naively, and if $b+d=N+1$, we insert $(a+c)/(b+d)$ in between $a/b$ and $c/d$. We have thus found an algorithm to generate any sequence $\mathfrak{F}_N$, for each $N$. Another interesting and key property of the sequence is that for any two consecutive fractions in any Farey sequence, say, $a/b$ and $c/d$, we have $ad-bc=-1$. Usually, this is called the determinant property. That’s great and all, but how is this related to our goal of approximating real numbers with relatively small denominators? We now turn to considering these fractions geometrically.

Fractions… but circles?

Let $a/b$ be a Farey fraction. We define $C(a/b)$ to be the circle that has centre at $(a/b,1/2b^2)$ with radius $1/2b^2$. These are known as Ford circles, named after Lester R Ford who studied them in 1938. Using our favourite computer program, we may draw out the Ford circles for the rationals in whatever $\mathfrak{F}_N$ we are interested in. Let’s stick to $N=9$. Listing out the elements of $\mathfrak{F}_9,$ we get
\begin{align*} \mathfrak{F}_9 = & \left( \frac{0}{1}, \frac{1}{9}, \frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{2}{9},\frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{4}{9}, \right. \\ & \left. \,\, \frac{1}{2}, \frac{5}{9}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{7}{9},\frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}, \frac{8}{9}, \frac{1}{1} \right). \end{align*}

Ford circles corresponding to Farey fractions with denominators $\leq 9$

Now, if we draw the Ford circles corresponding to each element in $\mathfrak{F}_9$, we have the image above.

Looking at the image, we may deduce a few properties: Each circle is tangent to the real axis at its corresponding fraction, and the circles for consecutive Farey fractions are tangent to each other! As a side note, by our algorithm to generate $\mathfrak{F}_N$, we may keep drawing more and more of these Ford circles over $[0,1]$, then use copies of them to extend the picture to the entire real line. To be more precise about the tangential circles phenomenon, one can establish the following result:

We will briefly sketch the proof of this. For the right implies left direction, what we really wish to establish is that the intersection of the interiors between distinct Ford circles is empty and that the circles touch. We can check this by investigating whether $$ \sqrt{\left(\frac{a}{b}-\frac{c}{d}\right)^2+\left(\frac{1}{2b^2}-\frac{1}{2d^2}\right)^2} \geq \frac{1}{2b^2}+\frac{1}{2d^2}. $$

Notice that all this inequality is really saying is that the distance between the centres of any two distinct Ford circles $C(a/b)$ and $C(c/d)$ is at least the sum of their radii. If we square both sides and simplify it a bit, the inequality above is equivalent to $$ (ad-bc)^2 \geq 1. $$ This holds for all rational numbers $a/b$ and $c/d$ (as long as $a/b$ and $c/d$ are distinct from each other — and we already decided they were, so that’s fine). Now if we specify that $a/b$ and $c/d$ are consecutive elements of $\mathfrak{F}_N$ for some $N$, we know from before that $ad-bc=-1$, and hence the inequality is sharp! In other words, recalling the inequality we began with, the distance between the two centres of the Ford circles $C(a/b)$ and $C(c/d)$ is exactly the sum of their radii, which implies that the circles are tangential to each other. The other direction isn’t too hard either, but it requires a few technical lemmas, so let’s move on…

Now, to finally answer our question, can we get a better bound? Yes! First, we assume without loss of generality that $\alpha$ is an irrational number between $0$ and $1$. If $\alpha >1 $ we can always add multiples of $q$ in the numerator of our rational approximation (ie the $p$ that appears in our inequality in the previous section), and we are supposing $\alpha$ to be irrational, since unsurprisingly approximating rational numbers with rational numbers is quite boring. Now let’s draw a vertical ray starting from $\alpha$. If we drew the Ford circles for all $N\in \mathbb{N}$ (whatever that means), this ray would intersect with infinitely many of them. Suppose that this ray intersects with $C(a,b)$, the Ford circle corresponding to the fraction $a/b$. By construction, our circle has radius $1/2b^2$.

A real number $\alpha$ as a vertical ray with our Ford circles

Let us focus on the $\alpha$ ray and the Ford circle $C(5/7)$ for now. After looking at the picture below, hopefully you will be convinced when I claim that the difference between our real number $\alpha$ and the Farey fraction $5/7$, or in this case actually, our rational approximation to $\alpha$ is bounded above by the radius of the circle, which in this case would be $1/98$.

Zoomed-in image of the Ford Circle $C(5/7)$ with the real number $\alpha$ ray

More generally, for any arbitrary real number $\alpha$, we may find a rational approximation in the following manner: we consider the Farey fractions $a/b$ of $\mathfrak{F}_N$, for whatever $N$ we wish. The punchline is that since the radius of each Ford circle is by construction $1/2b^2$, the distance between $\alpha$ and our rational approximation will always be bounded above by $1/2b^2$. We have now successfully obtained a better bound! So, we have improved our initial result from $1/2q$ to $1/2q^2$. Therefore, for any real $\alpha$ there exists some rational $p/q$ in lowest terms such that $$ \left| \alpha – \frac{p}{q} \right| \leq \frac{1}{2q^2}. $$ As promised, this ill-defined addition of rationals does lead to some interesting maths! In the odd event that you are not fully convinced yet, let’s consider the cardinality of each $\mathfrak{F}_N$. Here are some numerics: $$
\begin{gathered} \#\mathfrak{F}_1 =2, \quad \#\mathfrak{F}_2 = 3, \quad \#\mathfrak{F}_3 = 5, \quad \#\mathfrak{F}_4 = 7, \\ \#\mathfrak{F}_5 = 11, \quad \#\mathfrak{F}_6 = 13, \quad \dots \end{gathered} $$

They all seem to be primes. What possible connection can adding fractions ‘incorrectly’ have with the primes!? Or being even more naive/ambitious: is this at all related to the Riemann hypothesis? It turns out that the answer to that is yes. The Farey sequences, in particular their size and how the Farey fractions are distributed, are related to the Riemann hypothesis. The slightly underwhelming part is that it probably isn’t in the way you expected it to be. If we were to continue computing the size of the $\mathfrak{F}_N$s, disappointingly, $\#\mathfrak{F}_{10}=33,$ which isn’t prime…

Maybe… the Riemann hypothesis?

Before moving on, we will first introduce Big-O notation. Suppose $f(x)$ and $g(x)$ are two functions defined on some subset of the real numbers. We write $f(x)=O(g(x))$ if and only if there exists a constant $L$ and a positive constant $M$ such that $|f(x)| \leq M|g(x)|$ for all $x>L$. If you prefer plain English, just as I do, this can be thought of as ‘$f$ is bounded above by $g$, up to a constant, for sufficiently large $x$’. Now, tweaking our definition of $\mathfrak{F}_N$ slightly, we exclude the fraction $0/1$. Introducing some more notation, we let $\mathfrak{F}_N (i )$ denote the $i$th element of $\mathfrak{F}_N$, and let $\#\mathfrak{F}_N=m$. As an example, $\mathfrak{F}_3(1)=1/3$, and $\mathfrak{F}_4 (5)=3/4$. In 1924, Edmund Landau, building on previous work by Jérôme Franel, showed that the claim that for all $\varepsilon>0$, $$ S(N) = \sum_{i=1}^m \left|\mathfrak{F}_N(i) – \frac{i}{m} \right| = O\left(N^{1/2+\varepsilon}\right) $$ is equivalent to the Riemann hypothesis. The summand in question is the distance between the $i$th element of $\mathfrak{F}_N$ and the fraction $i/m$. In other words, heuristically, when $N$ is large, the Farey fractions of order $N$ are, roughly speaking, uniformly distributed, since the sum of these distances cannot grow too large. Here are some numerical values:

Even as $N$ and $m$ get larger, $S(N)$ stays relatively small!

Hopefully, by now, you are convinced that this ‘wrong way’ of adding fractions is not entirely dubious. Maybe someday you will prove the Riemann hypothesis via this equivalent statement, and if that’s the case, maybe let me squeeze my name onto your paper…

Wayne Junn Voo

Wayne is a mathematics undergraduate at the University of Warwick. He thinks about doing maths more than actually doing maths… Wayne would like to express his utmost gratitude to Arshay Sheth for providing valuable feedback on early versions of the article and, more importantly, for making all of this possible.

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