In my previous note, we considered equidistribution of rational points on the circle $X^2 + Y^2 = 2$. This is but one of a large family of equidistribution results + that + I'm not particularly familiar with.

This note is the first in a series of notes dedicated to exploring this type of equidistribution visually. In this note, we will investigate a simpler case — rational points on the line.

We know that $\mathbb{Q}$ is dense in $\mathbb{R}$. An equidistribution statement is roughly a way of quantifying this density. Let $I \subseteq \mathbb{R}$ be a finite interval. We will say that a sequence ${x_n}_{n \geq 1}$ of elements $x_n \in I$ is $\mu$-equidistributed on $I$ (or equidistributed with respect to $\mu$) if for each subinterval $J \subset I$ we have that $$\lim_{X \to \infty} \frac{\#\{n \leq X : x_n \in J\}}{X} = \int_J \mu(x) dx.$$ If $I = (\alpha, \beta)$ and $\mu = 1/(\beta - \alpha)$, then we will simply say that the sequence ${x_n}$ is equidistributed in $I$.

Note that $\mu$-equidistribution of a sequence implies that the sequence is dense in $I$, but the converse is not true. Further, ordering within the sequence matters — it is possible (and indeed, not hard at all) to reorder an equidistributed sequence ${ x_n }$ on $[0, 1]$ into a sequence ${ y_n }$ that is not equidistributed on $[0, 1]$. (For example, interlace rationals in $[0, 1/2]$ sparsely among the rationals $[1/2, 1]$, ordered by denominator size).

In this note, we will consider $\mathbb{Q}$. There are many enumerations of $\mathbb{Q}$ — but which enumeration will form our sequence? We will consider two such enumerations.

Let $q$ be a rational, written as $q = a/b$ (written in least terms). Define
$h_1(q) = b$, and define $h_2(q) = \sqrt{a^2 + b^2}$. These are two
*height* functions, giving some notion of the complexity of a number.
The first, $h_1$, says that the complexity of a rational is roughly the size of
the denominator. $h_2$ says that the complexity of a rational depends on both
the numerator and denominator, and numbers with a large numerator or
denominator are more complex.

For an implicit (finite) inverval $I$, let $X_1$ be an enumeration of rationals $q \in I$, ordered by $h_1$, and let $X_2$ be an enumeration ordered by $h_2$. We will consider equidistribution statements of the form $$\lim_{X \to \infty} \frac{\#\{x_n \in J: h_i(x_n) \leq X\}}{\#\{x_n \in I : h_i(x_n) \leq X \}} = \int_J \mu(x) dx.$$ (This makes the ambiguous ordering of elements with the same height unimportant).

It turns out that both $X_1$ and $X_2$ are sequences with some sense of
equidistribution, **but they are equidistributed with respect to
different functions.** The sequence $X_1$ is simply equidistributed (and
this can be proved using little more than the pigeonhole principle). The
sequence $X_2$ is equistributed with respect to $$ \mu(x) = \frac{1}{\pi}
\frac{1}{1 + t^2}. $$ I do not find this result obvious at all. I only learned
this fact recently, and I do not reproduce a proof here.

## Visualizing these rational points

Let us now turn to visualizing rational points in the two sequences $X_1$ and
$X_2$. Given a rational point $q$, we will visualize the data pair $(q, h(q))$,
where $h$ is a height function. Thus the *higher* that a point is within
the visualization, the higher the height.

These are images formed from all rationals on $[0, 3]$ with denominator bounded by $400$ (first image) and $100$ (second image). Each point corresponds to a point $(q, h_1(q))$.

These are images formed from all rationals $q$ on $[0, 3]$ with $h_2(q)^2 \leq 70000$ (first image) and $20000$ (second image). Each point corresponds to a point $(q, h_2(q))$.

Several patterns emerge from the images. At the bottom of each graph, there are
*wells* in which no points occur. If you examine these wells, you can
see that at the bottom center of each well, there is a single rational point.
These notably occur around points of low height — most notably around $0,
1/2, 1, 3/2, 2, 5/2, 3$. The reason is simple: for a rational $q = a/b$ to be
near $1/2$ (say), we need $1/2 - a/b \sim 1/2b$ to be small, and thus $b$ to be
large. In both $h_1$ and $h_2$, the height grows linearly in the denominator
$b$. On the other hand, there will be many other pairs of points of similarly
bounded height that are much closer. For example, $1/b$ and $1/(b-1)$ are
approximately $1/b^2$ apart, which can be significantly closer.^{1}
^{1}My
mathematical sibling Alex Walker first described this heuristic to me.

The graphs of $X_1$ visually become uniform, and appear almost to be mildly textured swaths of grey. We can make this even more pronounced, which we do in the next image. Graphs of $X_2$ show a clearly denser set of points at the left (favoring smaller numerators).

We now make a set of related images. In the images that follow, we extend each line from a point $(q, h(q))$ upwards. The effect is that at any designated height $H$, the horizontal line through $h(q) = H$ will include points whose heights are bounded above by $H$. To indicate the density of lines, we use shades of grey. The darker the line, the more rational points.

For $X1$, we make two images. First we have a *full* image, consisting
of points with $h(q) \leq 400$ and $20$ different shades of grey. Second, we
have a *sparser* image, consisting of points with $h(q) \leq 100$ and
$12$ different shades of grey. These are still over the interval $[0, 3]$.

For $X2$, we first have those points with $h(q) \leq 70000$ and $16$ different shades, and then an image with $h(q) \leq 10000$ and $10$ different shades.

These images really emphasize the *wells* around rational points of low
height. These gives these images their texture.

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