Vanishing derivative in rational points

[Tolaso J Kos]

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f'(x)=0, \;\; \forall x\in \mathbb{Q}$. Prove that $f$ is constant in $\mathbb{R}$.

[Grigorios Kostakos]

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f'(x)=0, \;\; \forall x\in \mathbb{Q}$. Prove that f is constant in $\mathbb{R}$.

Because \(f:\mathbb{R}\longrightarrow\mathbb{R}\) is differentiable in \(\mathbb{R}\) with $f'(x)=0\,,\quad \forall\,x\in\mathbb{Q}$, we have that \(f\) is continuous in \(\mathbb{R}\) and $f(x)=c\,,\quad \forall\,x\in\mathbb{Q}$. Let \(x_0\in\mathbb{R}\setminus\mathbb{Q}\) and assume that \(f(x_0)\neq c\). Because \(\mathbb{Q}\) is dense subset of \(\mathbb{R}\), there is a sequence \(\bigl\{{q_n}\bigr\}_{n=1}^{\infty}\) of rationals which converges to \(x_0\). So we must have \[c\neq f(x_0)=\mathop{\lim}\limits_{x\to x_0}{f(x)}=\mathop{\lim}\limits_{n\to +\infty}{f(q_n)}=\mathop{\lim}\limits_{n\to +\infty}{c}=c\] A contradiction. So \[f(x)=c\,,\quad \forall\,x\in\mathbb{R}\,.\]