The Dirichlet function $1_{\mathbb{Q}}$, and Thomae's function $t$ with $t(x)=1\mathrm{/}q$ at a rational $x=p\mathrm{/}q$ in lowest terms with $q\ge 1$ and $t(x)=0$ at every irrational $x$

Throughout, $\mathbb{N}\subseteq \mathbb{Z}\subseteq \mathbb{Q}\subseteq \mathbb{R}$ denotes the chain of canonical embeddings (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals), and each set is identified with its image in $\mathbb{R}$, as elsewhere in this library. A natural $q$ denotes its canonical real value $q\cdot 1_{\mathbb{R}}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), and $q>0$ for $q\ge 1$ (Canonical naturals are positive and strictly increasing). A real is rational when it lies in $\mathbb{Q}$ and irrational otherwise; both sets are dense in $\mathbb{R}$ (Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable).

The Dirichlet function

${\mathbf{1}}_{\mathbb{Q}}:\mathbb{R}\to \mathbb{R},{\mathbf{1}}_{\mathbb{Q}}(x):=\{\begin{array}{ll}1& \text{if }x\in \mathbb{Q},\\ 0& \text{if }x\mathrm{\notin }\mathbb{Q}\mathrm{.}\end{array}$

This is the indicator of the rationals. It is a function because every real either lies in $\mathbb{Q}$ or does not, and the two clauses are exclusive.

The least denominator of a rational

Let $x\in \mathbb{Q}$ and put

$Q(x):=\{q\in \mathbb{N}:q\ge 1\text{ and }qx\in \mathbb{Z}\}\mathrm{.}$

$Q(x)$ is nonempty. Every rational is $a\mathrm{/}b$ with $a\in \mathbb{Z}$ and $b$ a positive integer (The integers embed in the rationals), and a positive integer is $q$ for a unique natural $q\ge 1$ (The naturals embed in the integers); then $qx=a\in \mathbb{Z}$, so $q\in Q(x)$.

By the well-ordering principle (The well-ordering principle) the nonempty subset $Q(x)\subseteq \mathbb{N}$ has a least element. Write

$q(x):=\min Q(x)\ge 1,$

the least denominator of $x$, and $p(x):=q(x)x\in \mathbb{Z}$, so that

$x=\frac{p(x)}{q(x)}\mathrm{.}$

Nothing is selected here: $q(x)$ is the least element of a set determined by $x$, so it is a function of $x$ alone.

The least denominator is the denominator in lowest terms. The integers $p(x)$ and $q(x)$ are coprime (Coprime integers: $\gcd (a,b)=1$). Indeed put $d:=\gcd (p(x),q(x))$, which satisfies $d\ge 1$ because $q(x)\ge 1$ makes the pair different from $(0,0)$ ($\gcd$ is symmetric and unchanged by signs: $\gcd (a,b)=\gcd (b,a)=\gcd (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$; moreover $\gcd (a,0)=\mathrm{\mid }a\mathrm{\mid }$, $\gcd (a,1)=1$, $\gcd (a,a)=\mathrm{\mid }a\mathrm{\mid }$, and $\gcd (a,b)\ge 1$ unless $a=b=0$, Common divisor, and the greatest common divisor $\gcd (a,b)$, with the convention $\gcd (0,0):=0$). Then $d$ divides $q(x)$, so $q(x)\mathrm{/}d$ is a natural number $\ge 1$. Since the canonical embeddings preserve products, this integer quotient agrees with real division (Canonical naturals are positive and strictly increasing); hence

$(q(x)\mathrm{/}d)x=\frac{q(x)x}{d}=\frac{p(x)}{d}\in \mathbb{Z},$

so $q(x)\mathrm{/}d\in Q(x)$ and therefore $q(x)\mathrm{/}d\ge q(x)$, which forces $d=1$. Conversely, a lowest-terms denominator is the least one, so the description is unambiguous. Suppose $x=p\mathrm{/}q$ with $q\ge 1$ a natural, $p\in \mathbb{Z}$ and $\gcd (p,q)=1$. Then $q\in Q(x)$, so $q_0:=q(x)\le q$; and from $p\mathrm{/}q=p(x)\mathrm{/}{q}_0$ we get $q_{0}p=qp(x)$ in $\mathbb{Z}$, so $q\mid q_{0}p$, and $\gcd (p,q)=1$ gives $q\mid q_0$ (If $\gcd (a,b)=1$ and $a\mid bc$ then $a\mid c$; and if $a\mid c$, $b\mid c$ and $\gcd (a,b)=1$ then $ab\mid c$, claim 1), hence $q\le q_0$. So $q=q(x)$: writing $x=p\mathrm{/}q$ "in lowest terms with $q\ge 1$" and taking $q=q(x)$ describe the same integer, and If $d=\gcd (a,b)$ is nonzero then $a\mathrm{/}d$ and $b\mathrm{/}d$ are coprime is what produces such a representation from an arbitrary one.

Thomae's function

$t:\mathbb{R}\to \mathbb{R},t(x):=\{\begin{array}{ll}1\mathrm{/}q(x)& \text{if }x\in \mathbb{Q},\\ 0& \text{if }x\mathrm{\notin }\mathbb{Q}\mathrm{.}\end{array}$

It is also called the popcorn function or the ruler function. The value is well defined because $q(x)$ is, and $q(x)\ge 1>0$ is invertible.

Boundary values, stated rather than left to the reader.

• $t(0)=1$. Indeed $1\cdot 0=0\in \mathbb{Z}$, so $1\in Q(0)$ and $q(0)=1$; the representation is $0=0\mathrm{/}1$.
• $t(m)=1$ for every integer $m$, by the same computation with $1\in Q(m)$.
• $0<t(x)\le 1$ for every rational $x$, since $q(x)\ge 1$; and $t(x)=0$ exactly at the irrationals.

On the range. The values of $t$ are $0$ and the reciprocals $1\mathrm{/}q$ of the canonical naturals $q\ge 1$; every such value is attained, $1\mathrm{/}q$ being the value at the rational $1\mathrm{/}q$ itself, whose least denominator is $q$ because $k\mathrm{/}q\in \mathbb{Z}$ with $1\le k<q$ would give a positive integer smaller than $1$.