# What does the graph of the height function on $\mathbf{Q}$ look like

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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It's not clear to me what you're asking. Do you mean literally what it would look like? It wouldn't look like anything because as you write most of it would be beyond any bound and hence not visible to the naked eye. What other sense of "look like" do you mean that doesn't coincide with "what values does it take" (which you already know)? –  joriki Nov 11 '11 at 22:34
The function $q\mapsto \frac{1}{h(q)}$ would probably look more interesting. See also Thomae's function for a slightly related function. –  Henning Makholm Nov 11 '11 at 23:42
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.