* For rational numbers, this follows from non-continuity.
* For irrational numbers:
*:For any [[sequence]] of irrational numbers <math>(a_n)_{n=1}^\infty</math> with <math>a_n \ne x_0</math> for all <math>n \in \mathbb{N}_{+}</math> that converges to the irrational point <math>x_0,\;</math>, the sequence <math>(f(a_n))_{n=1}^\infty</math> is identically <math>0,\;</math>, and so <math>\lim_{n \to \infty}\left|\frac{f(a_n)-f(x_0)}{a_n - x_0}\right| = 0.</math>.
*:AccordingtoOn [[Hurwitz'sthe theoremother (number theory)|Hurwitz's theorem]]hand, thereconsider also exists athe sequence of rational numbers <math>(b_n)_{n=1}^{\infty}</math> =with <math>b_n (k_n/n)_{n=1}^ \infty,lfloor nx_0\;rfloor/n</math>, converging towhere <math>x_0,\;lfloor nx_0\rfloor</math> withdenotes the [[Floor and ceiling functions|floor]] of <math>k_nnx_0</math>. Since <math>nx_0-1<\inlfloor nx_0\rfloor\mathbble Znx_0</math>, andthe sequence <math>(b_n)_{n=1}^{\ininfty}</math> \mathbbconverges Nto <math>x_0</math> coprimeusing andthe [[Squeeze theorem]]. Also, <math>|k_nb_n-x_0| = |\lfloor nx_0\rfloor/n - x_0| <= |\frac{1}{lfloor nx_0\sqrt{5}\cdotrfloor - nx_0|/n^2}. \;le 1/n</math> for all <math>n</math>.
*: Thus for all <math>n,</math>, <math>\left| \frac{f(b_n)-f(x_0)}{b_n-x_0} \right| >\ge \frac{1/n - 0}{1/(n} = 1</math>. Therefore we obtain <math>\sqrtliminf_{5n\to\infty} \cdotleft| n^2\frac{f(b_n)-f(x_0)} =\sqrt{5b_n-x_0} \cdotright| n\ge 1 \ne 0\;</math> and so {{nowrap|<math>f</math> is not differentiable}} at allany irrational number <math>x_0.</math>.
