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

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

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

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

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

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

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

Let ${\displaystyle \;\varepsilon =1/q\;}$ , and given ${\displaystyle \delta >0}$  let ${\displaystyle n=1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil .}$  For the corresponding ${\displaystyle \;x_{n}}$  we have

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

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

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

The minimal distance of ${\displaystyle x_{0}}$  to its i-th lower and upper bounds equals

We define ${\displaystyle \delta }$  as the minimum of all the finitely many ${\displaystyle d_{i}.}$ 

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

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

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

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

• For rational numbers, this follows from non-continuity.
• For irrational numbers:

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

  According to Hurwitz's theorem, there also exists a sequence of rational numbers ${\displaystyle (b_{n})_{n=1}^{\infty }=(k_{n}/n)_{n=1}^{\infty },\;}$  converging to ${\displaystyle x_{0},\;}$  with ${\displaystyle k_{n}\in \mathbb {Z} }$  and ${\displaystyle n\in \mathbb {N} }$  coprime and ${\displaystyle |k_{n}/n-x_{0}|<{\frac {1}{{\sqrt {5}}\cdot n^{2}}}.\;}$ 

  Thus for all ${\displaystyle n,}$  ${\displaystyle \left|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right|>{\frac {1/n-0}{1/({\sqrt {5}}\cdot n^{2})}}={\sqrt {5}}\cdot n\neq 0\;}$  and so ${\displaystyle f}$  is not differentiable at all irrational ${\displaystyle x_{0}.}$