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This question is a bit vague. I was just wondering what the graph of the height function on $\mathbf{Q}$ would look like.

Define the height $h(q)$ of a rational number $q$ as follows. Write $q=a/b$, where $a$ and $b$ are coprime integers. Then $h(q) := \max(\vert a\vert,\vert b\vert)$.

The function $h$ has the property that the set of rational numbers $q$ such that $h(q)$ is bounded by a real number $C$ is finite. So I was trying to imagine how this would look like on the interval $[0,1]$ but didn't get really far. It gets arbitrarily large around any number $x \in [0,1]$.

Any thoughts?

Of course, we could also consider a height function on $\overline{\mathbf{Q}}$ and its graph.

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    The function $q\mapsto \frac{1}{h(q)}$ would probably look more interesting. See also [Thomae's function](http://en.wikipedia.org/wiki/Thomae%27s_function) for a slightly related function.2011-11-11

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First, the graph is symmetric about $0$. It also satisfies "inversion:" $h(q) = h(\frac{1}{q})$, so in a sense you only need to know what it looks like on $(0,1]$ or on $[1,\infty)$. On $[1,\infty)$, you have one point at level $1$ (the point $(1,1)$); then one point on level $2$, $(2,2)$; then two points on level $3$: $(3,3)$, $(3/2,3)$. Then two points on level $4$, namely $(4,4)$ and $(4/3,4)$.

In general, at level $n$ you have the $\varphi(n)$ points $(\frac{n}{k},n)$, where $1\leq k\lt n$ and $\gcd(k,n)=1$.

Now "invert" to get the graph on $(0,1]$; and reflect about the $y$ axis to get the whole graph.