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Thomae's function

Function that is discontinuous at rationals and continuous at irrationals

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Source updated August 16, 2026

Thomae's function is a real-valued function of a real variable that can be defined as: $f(x)={\begin{cases}{\frac {1}{q}}&{\text{if }}x={\tfrac {p}{q}}\quad (x{\text{ is rational), with }}p\in \mathbb {Z} {\text{ and }}q\in \mathbb {N} {\text{ coprime}}\\0&{\text{if }}x{\text{ is irrational.}}\end{cases}}$

It is named after Carl Johannes Thomae, but has many other names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function (not to be confused with the integer ruler function), the Riemann function, or the Stars over Babylon (John Horton Conway's name). Thomae mentioned it as an example for an integrable function with infinitely many discontinuities in an early textbook on Riemann's notion of integration.

Since every rational number has a unique representation with coprime (also termed relatively prime) $p\in \mathbb {Z}$ and $q\in \mathbb {N}$ , the function is well-defined. Note that $q=+1$ is the only number in $\mathbb {N}$ that is coprime to $p=0.$

It is a modification of the Dirichlet function, which is 1 at rational numbers and 0 elsewhere.

01Properties

Thomae's function $f$ is bounded and maps all real numbers to the unit interval: $f:\mathbb {R} \to [0,1].$

$f$ is periodic with period $1:\;f(x+n)=f(x)$ for all integers n and all real x.

Proof of periodicity
For all $x\in \mathbb {R} \setminus \mathbb {Q} ,$ we also have $x+n\in \mathbb {R} \setminus \mathbb {Q}$ and hence $f(x+n)=f(x)=0,$ For all $x\in \mathbb {Q} ,\;$ there exist $p\in \mathbb {Z}$ and $q\in \mathbb {N}$ such that $\;x=p/q,\;$ and $\gcd(p,\;q)=1.$ Consider $x+n=(p+nq)/q$ . If $d$ divides $p$ and $q$ , it divides $p+nq$ and $q$ . Conversely, if $d$ divides $p+nq$ and $q$ , it divides $(p+nq)-nq=p$ and $q$ . So $\gcd(p+nq,q)=\gcd(p,q)=1$ , and $f(x+n)=1/q=f(x)$ .

Proof of periodicity

For all $x\in \mathbb {R} \setminus \mathbb {Q} ,$ we also have $x+n\in \mathbb {R} \setminus \mathbb {Q}$ and hence $f(x+n)=f(x)=0,$

For all $x\in \mathbb {Q} ,\;$ there exist $p\in \mathbb {Z}$ and $q\in \mathbb {N}$ such that $\;x=p/q,\;$ and $\gcd(p,\;q)=1.$ Consider $x+n=(p+nq)/q$ . If $d$ divides $p$ and $q$ , it divides $p+nq$ and $q$ . Conversely, if $d$ divides $p+nq$ and $q$ , it divides $(p+nq)-nq=p$ and $q$ . So $\gcd(p+nq,q)=\gcd(p,q)=1$ , and $f(x+n)=1/q=f(x)$ .

$f$ is discontinuous at every rational number, so its points of discontinuity are dense within the real numbers.

Proof of discontinuity at rational numbers
Let $x_{0}=p/q$ be an arbitrary rational number, with $\;p\in \mathbb {Z} ,\;q\in \mathbb {N} ,$ and $p$ and $q$ coprime. This establishes $f(x_{0})=1/q.$ Let $\;\alpha \in \mathbb {R} \setminus \mathbb {Q} \;$ be any irrational number and define $x_{n}=x_{0}+{\frac {\alpha }{n}}$ for all $n\in \mathbb {N} .$ These $x_{n}$ are all irrational, and so $f(x_{n})=0$ for all $n\in \mathbb {N} .$ This implies $\|x_{0}-x_{n}\|={\frac {\alpha }{n}},$ and $\|f(x_{0})-f(x_{n})\|={\frac {1}{q}}.$ Let $\;\varepsilon =1/q\;$ , and given $\delta >0$ let $n=1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil .$ For the corresponding $\;x_{n}$ we have $\|f(x_{0})-f(x_{n})\|=1/q\geq \varepsilon$ and $\|x_{0}-x_{n}\|={\frac {\alpha }{n}}={\frac {\alpha }{1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}<{\frac {\alpha }{\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}\leq \delta ,$ which is exactly the definition of discontinuity of $f$ at $x_{0}$ .

Proof of discontinuity at rational numbers

Let $x_{0}=p/q$ be an arbitrary rational number, with $\;p\in \mathbb {Z} ,\;q\in \mathbb {N} ,$ and $p$ and $q$ coprime.

This establishes $f(x_{0})=1/q.$

Let $\;\alpha \in \mathbb {R} \setminus \mathbb {Q} \;$ be any irrational number and define $x_{n}=x_{0}+{\frac {\alpha }{n}}$ for all $n\in \mathbb {N} .$

These $x_{n}$ are all irrational, and so $f(x_{n})=0$ for all $n\in \mathbb {N} .$

This implies $|x_{0}-x_{n}|={\frac {\alpha }{n}},$ and $|f(x_{0})-f(x_{n})|={\frac {1}{q}}.$

Let $\;\varepsilon =1/q\;$ , and given $\delta >0$ let $n=1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil .$ For the corresponding $\;x_{n}$ we have $|f(x_{0})-f(x_{n})|=1/q\geq \varepsilon$ and $|x_{0}-x_{n}|={\frac {\alpha }{n}}={\frac {\alpha }{1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}<{\frac {\alpha }{\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}\leq \delta ,$

which is exactly the definition of discontinuity of $f$ at $x_{0}$ .

$f$ is continuous at every irrational number, so its points of continuity are dense within the real numbers.

Proof of continuity at irrational arguments
Since $f$ is periodic with period $1$ and $0\in \mathbb {Q} ,$ it suffices to check all irrational points in $I=(0,1).\;$ Assume now $\varepsilon >0,\;i\in \mathbb {N}$ and $x_{0}\in I\setminus \mathbb {Q} .$ According to the Archimedean property of the reals, there exists $r\in \mathbb {N}$ with $1/r<\varepsilon ,$ and there exist $\;k_{i}\in \mathbb {N} ,$ such that for $i=1,\ldots ,r$ we have $0\leq {\frac {k_{i}}{i}}<x_{0}<{\frac {k_{i}+1}{i}}.$ The minimal distance of $x_{0}$ to its i-th lower and upper bounds equals $d_{i}:=\min \left\{\left\|x_{0}-{\frac {k_{i}}{i}}\right\|,\;\left\|x_{0}-{\frac {k_{i}+1}{i}}\right\|\right\}.$ We define $\delta$ as the minimum of all the finitely many $d_{i}.$ $\delta :=\min _{1\leq i\leq r}\{d_{i}\},\;$ so that for all $i=1,\dots ,r,$ $\|x_{0}-k_{i}/i\|\geq \delta$ and $\|x_{0}-(k_{i}+1)/i\|\geq \delta .$ This is to say, all these rational numbers $k_{i}/i,\;(k_{i}+1)/i,\;$ are outside the $\delta$ -neighborhood of $x_{0}.$ Now let $x\in \mathbb {Q} \cap (x_{0}-\delta ,x_{0}+\delta )$ with the unique representation $x=p/q$ where $p,q\in \mathbb {N}$ are coprime. Then, necessarily, $q>r,\;$ and therefore, $f(x)=1/q<1/r<\varepsilon .$ Likewise, for all irrational $x\in I,\;f(x)=0=f(x_{0}),\;$ and thus, if $\varepsilon >0$ then any choice of (sufficiently small) $\delta >0$ gives $\|x-x_{0}\|<\delta \implies \|f(x_{0})-f(x)\|=f(x)<\varepsilon .$ Therefore, $f$ is continuous on $\mathbb {R} \setminus \mathbb {Q} .$

Proof of continuity at irrational arguments

Since $f$ is periodic with period $1$ and $0\in \mathbb {Q} ,$ it suffices to check all irrational points in $I=(0,1).\;$ Assume now $\varepsilon >0,\;i\in \mathbb {N}$ and $x_{0}\in I\setminus \mathbb {Q} .$ According to the Archimedean property of the reals, there exists $r\in \mathbb {N}$ with $1/r<\varepsilon ,$ and there exist $\;k_{i}\in \mathbb {N} ,$ such that

for $i=1,\ldots ,r$ we have $0\leq {\frac {k_{i}}{i}}<x_{0}<{\frac {k_{i}+1}{i}}.$

The minimal distance of $x_{0}$ to its i-th lower and upper bounds equals $d_{i}:=\min \left\{\left|x_{0}-{\frac {k_{i}}{i}}\right|,\;\left|x_{0}-{\frac {k_{i}+1}{i}}\right|\right\}.$

We define $\delta$ as the minimum of all the finitely many $d_{i}.$ $\delta :=\min _{1\leq i\leq r}\{d_{i}\},\;$ so that for all $i=1,\dots ,r,$ $|x_{0}-k_{i}/i|\geq \delta$ and $|x_{0}-(k_{i}+1)/i|\geq \delta .$

This is to say, all these rational numbers $k_{i}/i,\;(k_{i}+1)/i,\;$ are outside the $\delta$ -neighborhood of $x_{0}.$

Now let $x\in \mathbb {Q} \cap (x_{0}-\delta ,x_{0}+\delta )$ with the unique representation $x=p/q$ where $p,q\in \mathbb {N}$ are coprime. Then, necessarily, $q>r,\;$ and therefore, $f(x)=1/q<1/r<\varepsilon .$

Likewise, for all irrational $x\in I,\;f(x)=0=f(x_{0}),\;$ and thus, if $\varepsilon >0$ then any choice of (sufficiently small) $\delta >0$ gives $|x-x_{0}|<\delta \implies |f(x_{0})-f(x)|=f(x)<\varepsilon .$

Therefore, $f$ is continuous on $\mathbb {R} \setminus \mathbb {Q} .$

$f$ is nowhere differentiable.

Proof of being nowhere differentiable
For rational numbers, this follows from non-continuity. For irrational numbers: For any sequence of irrational numbers $(a_{n})_{n=1}^{\infty }$ with $a_{n}\neq x_{0}$ for all $n\in \mathbb {N} _{+}$ that converges to the irrational point $x_{0}$ , the sequence $(f(a_{n}))_{n=1}^{\infty }$ is identically $0$ , and so $\lim _{n\to \infty }\left\|{\frac {f(a_{n})-f(x_{0})}{a_{n}-x_{0}}}\right\|=0$ . On the other hand, consider the sequence of rational numbers $(b_{n})_{n=1}^{\infty }$ with $b_{n}=\lfloor nx_{0}\rfloor /n$ , where $\lfloor nx_{0}\rfloor$ denotes the floor of $nx_{0}$ . Since $nx_{0}-1<\lfloor nx_{0}\rfloor \leq nx_{0}$ , the sequence $(b_{n})_{n=1}^{\infty }$ converges to $x_{0}$ using the Squeeze theorem. Also, $\|b_{n}-x_{0}\|=\|\lfloor nx_{0}\rfloor /n-x_{0}\|=\|\lfloor nx_{0}\rfloor -nx_{0}\|/n\leq 1/n$ for all $n$ . Thus for all $n$ , $\left\|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right\|\geq {\frac {1/n-0}{1/n}}=1$ . Therefore we obtain $\liminf _{n\to \infty }\left\|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right\|\geq 1\neq 0$ and so $f$ is not differentiable at any irrational number $x_{0}$ .

Proof of being nowhere differentiable

For rational numbers, this follows from non-continuity.
For irrational numbers:
For any sequence of irrational numbers $(a_{n})_{n=1}^{\infty }$ with $a_{n}\neq x_{0}$ for all $n\in \mathbb {N} _{+}$ that converges to the irrational point $x_{0}$ , the sequence $(f(a_{n}))_{n=1}^{\infty }$ is identically $0$ , and so $\lim _{n\to \infty }\left|{\frac {f(a_{n})-f(x_{0})}{a_{n}-x_{0}}}\right|=0$ .

On the other hand, consider the sequence of rational numbers $(b_{n})_{n=1}^{\infty }$ with $b_{n}=\lfloor nx_{0}\rfloor /n$ , where $\lfloor nx_{0}\rfloor$ denotes the floor of $nx_{0}$ . Since $nx_{0}-1<\lfloor nx_{0}\rfloor \leq nx_{0}$ , the sequence $(b_{n})_{n=1}^{\infty }$ converges to $x_{0}$ using the Squeeze theorem. Also, $|b_{n}-x_{0}|=|\lfloor nx_{0}\rfloor /n-x_{0}|=|\lfloor nx_{0}\rfloor -nx_{0}|/n\leq 1/n$ for all $n$ .

Thus for all $n$ , $\left|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right|\geq {\frac {1/n-0}{1/n}}=1$ . Therefore we obtain $\liminf _{n\to \infty }\left|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right|\geq 1\neq 0$ and so $f$ is not differentiable at any irrational number $x_{0}$ .

$f$ has a proper local maximum at each rational number, providing an example of a function with a dense set of proper local maxima.

See the proofs for continuity and discontinuity above for the construction of appropriate neighbourhoods, where $f$ has maxima.
$f$ is Riemann integrable on any interval and the integral evaluates to $0$ over any set.

The Lebesgue criterion for integrability states that a bounded function is Riemann integrable if and only if the set of all discontinuities has measure zero. Every countable subset of the real numbers - such as the rational numbers - has measure zero, so the above discussion shows that Thomae's function is Riemann integrable on any interval. The function's integral is equal to $0$ over any set because the function is equal to zero almost everywhere.
If $G=\{\,(x,f(x)):x\in (0,1)\,\}\subset \mathbb {R} ^{2}$ is the graph of the restriction of $f$ to $(0,1)$ , then the box-counting dimension of $G$ is $4/3$ .

03The ruler function

For integers, the exponent of the highest power of 2 dividing $n$ gives 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, ... (sequence A007814 in the OEIS). If 1 is added, or if the 0s are removed, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, ... (sequence A001511 in the OEIS). The values resemble tick-marks on a 1/16th graduated ruler, hence the name. These values correspond to the restriction of the Thomae function to the dyadic rationals: those rational numbers whose denominators are powers of 2.

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Sources and credits

This article is adapted from the Wikipedia article “Thomae's function”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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Thomae function (0,1).svg by Smithers888, Public domain

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Thomae's function

01Properties

03The ruler function

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Thomae's function

01Properties

02Related probability distributions

03The ruler function

04Related functions

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